Mathematics M.C.Q (1 Marks)

$\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2} \& \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$

Formula for shortest distance

S.D. $=\frac{\left|\begin{array}{ccc} x _2- x _1 & y_2-y_1 & z_2-z_1 \\ a _1 & b _1 & c _1 \\ a _2 & b _2 & c _2\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a _1 & b _1 & c _1 \\ a _2 & b _2 & c _2\end{array}\right|}$

$=\frac{\left|\begin{array}{ccc}6 & 1 & -8 \\ 1 & -2 & 2 \\ 1 & 2 & 0\end{array}\right|}{\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -2 & 2 \\ 1 & 2 & 0\end{array}\right|}=\frac{54}{6}=9$

$a \hat{i}+b \hat{j}+c \hat{k}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1\end{array}\right|$

$=\hat{ i }(2-6)-\hat{ j }(1-6)+\hat{ k }(2-4)$

$=-4 \hat{ i }-5 \hat{ j }-2 \hat{ k }$

$\ell=\gamma(-4 \hat{ i }+5 \hat{ j }-2 \hat{ k })$

$P$ is intersection of $\ell$ and $\ell_1$

$-4 \gamma=1+\lambda, 5 \gamma=-11+2 \lambda,-2 \gamma=-7+3 \lambda$

By solving there equation $\gamma=-1, P (4,-5,2)$

Let $Q (-1+2 \mu, 2 \mu, 1+\mu)$

$\overline{ PQ } \cdot(2 \hat{ i }+2 \hat{ j }+\hat{ k })=0$

$-2+4 \mu+4 \mu+1+\mu=0$

$9 \mu=1$

$\mu=\frac{1}{9}$

$Q\left(\frac{-7}{9}, \frac{2}{9}, \frac{10}{9}\right)$

$9(\alpha+\beta+\gamma)=9\left(\frac{-7}{9}+\frac{2}{9}+\frac{10}{9}\right)$

$=5$

$3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0$

$9-4 \mu=0$

$\mu=\frac{9}{4}$

$4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0$

$-100+4 \alpha-54+18 \alpha=0$

$\Rightarrow \alpha=7$

$\text { Let } P \equiv(3 \lambda-5, \lambda-4,-2 \lambda+7)$

$\text { Direction ratio of } P Q(3 \lambda-1, \lambda-1,-2 \lambda+5)$

$\text { But } P Q \perp \ell_1$

$\Rightarrow 3(3 \lambda-1)+1 \cdot(\lambda-1)-2(-2 \lambda+5)=0$

$\Rightarrow \lambda=1$

$P(-2,-3,5) \Rightarrow|a|+|b|+|c|=10$

$\sin 2 A+\sin 2 B+\sin 2 C=\frac{9}{x}$

$\therefore \sin A +\sin B+\sin C=2(\sin 2 A+\sin 2 B+\sin 2 C)$

$4 \cos A / 2 \cos B / 2 \cos C / 2=2(4 \sin A \sin B \sin C)$

$16 \sin A / 2 \sin B / 2 \sin C / 2=1$

$\text { Hence Ans. }=5$

And $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$ intersect

Point on first line $(1,2,3)$ and point on second line $(4,1,0)$.

Vector joining both points is $-3 \hat{i}+\hat{j}+3 \hat{k}$

Now vector along first line is $2 \hat{i}+3 \hat{j}+\alpha \hat{k}$

Also vector along second line is $5 \hat{i}+2 \hat{j}+\beta \hat{k}$

Now these three vectors must be coplanar

$\begin{aligned}& \Rightarrow\left|\begin{array}{ccc}2 & 3 & \alpha \\5 & 2 & \beta \\ -3 & 1 & 3\end{array}\right|=0 \\& \Rightarrow 2(6-\beta)-3(15+3 \beta)+\alpha(11)=0 \\ & \Rightarrow \alpha-\beta=3\end{aligned}$

Now $\alpha=3+\beta$

Given expression $8(3+\beta) \cdot \beta=8\left(\beta^2+3 \beta\right)$

$=8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right)=8\left(\beta+\frac{3}{2}\right)^2-18$

So magnitude of minimum value $=18$

$a _{1}=(1,2,3)$

$a _{2}=(2,4,5)$

$\overrightarrow{ b }_{2}=2 \hat{ i }+3 \hat{ j }+\lambda \hat{ k }$

$\overrightarrow{ b }_{2}=\hat{ i }+4 \hat{ j }+5 \hat{ k }$

S.D. $=\frac{\left|((2-1) \hat{i}+(4-2) \hat{j}+(5-3) \hat{k}) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}{\left|b_{1} \times b_{2}\right|}$

$\overrightarrow{ b }_{1} \times \overrightarrow{ b }_{2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|$

$=\hat{ i }(15-4 \lambda)+\hat{ j }(\lambda-10)+\hat{ k }(5)$

$=(15-4 \lambda) \hat{ i }+(\lambda-10) \hat{ j }+5 \hat{ k }$

$\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+25}$

S.D. $=\frac{|(\hat{ i }+2 \hat{ j }+2 \hat{ k }) \cdot[(15-4 \lambda) \hat{ i }+(\lambda-10) \hat{ j }+5 \hat{ k }]|}{\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+25}}$

$\frac{|15-4 \lambda+2 \lambda-20+10|}{\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+25}}=\frac{1}{\sqrt{3}}$

$3(5-2 \lambda)^{2}=225+16 \lambda^{2}-120 \lambda+\lambda^{2}+100-20 \lambda+25$

$12 \lambda^{2}+75-60 \lambda=17 \lambda^{2}-140 \lambda+350$

$5 \lambda^{2}-80 \lambda+275=0$

$\lambda^{2}-16 \lambda+55=0$

$(\lambda-5)(\lambda-11)=0$

$\Rightarrow \lambda=5,11$

$a_{2}=(0,-1,2)$

$b_{1}=(1,-a, 0)$ dr's of line ($1$)

$b_{2}=(1,-1,1)$ dr's of line $(2)$

$\overline{ a }_{2}-\overline{ a }_{1}=(1,-1,-1)$

$\overline{ b }_{1} \times \overline{ b }_{2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & - a & 0 \\ 1 & -1 & 1\end{array}\right|$

$\overline{ b }_{1} \times \overline{ b }_{2}=\hat{ i }(- a )-\hat{ j }+\hat{ k }( a -1)$

$\left|\bar{b}_{1} \times \bar{b}_{2}\right|=\sqrt{ a ^{2}+1+( a -1)^{2}}$

$a _{2}- a _{1} \cdot \overline{ b }_{1} \times \overline{ b }_{2}=2-2 a$

$\frac{2(1-a)}{\sqrt{a^{2}+1+(a-1)^{2}}}=\sqrt{\frac{2}{3}}$

Squaring an both the side

After solving $a=2, \frac{1}{2}$

$l_{2}: \frac{x-1}{1}=\frac{y+3 / 2}{\alpha / 2}=\frac{z+5}{2}$

$l_{3}: \frac{x-1}{-3}=\frac{y-1 / 2}{-2}=\frac{z-0}{4}$

$l_{1} \perp l_{2} \Rightarrow \frac{|3-\alpha+0|}{\sqrt{13} \sqrt{1+\frac{\alpha^{2}}{4}+4}}=0 \Rightarrow \alpha=3$

angle between $l_{2}$ and $l_{3}$

$\cos \theta=\frac{|1 \times(-3)+(-2)(\alpha / 2)+2 \times 4|}{\sqrt{1+4+\frac{\alpha^{2}}{4}} \sqrt{9+16+4}}$

$\cos \theta=\frac{|-3-\alpha+8|}{\sqrt{5+\frac{\alpha^{2}}{4}} \sqrt{29}}$

put $\alpha=3$

$\cos \theta=\frac{2}{\sqrt{\frac{29}{4} \sqrt{29}}}=\frac{4}{29}$

$\theta=\cos ^{-1}\left(\frac{4}{29}\right) \Rightarrow \theta=\sec ^{-1}\left(\frac{29}{4}\right)$

$\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$

$A =(3,2,1) \quad B =(-3,6,5)$

$\overline{n_{1}}=2 \hat{i}+3 \hat{j}-\hat{k}$

$\overrightarrow{ n _{2}}=2 \hat{ i }+\hat{ j }-3 \hat{ k }$

$\overrightarrow{ BA }=6 \hat{ i }-4 \hat{ j }-4 \hat{ k }$

Shortest Distance $=\frac{\left[\overrightarrow{ BA } \overline{n_{1}} \overrightarrow{n_{2}}\right]}{\left|\overline{n_{1}} \times \overline{n_{2}}\right|}$

$\overline{n_{1}} \times \overrightarrow{n_{2}}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|$

$=10 \hat{ i }-8 \hat{ j }-4 \hat{ k }$

$\left[\begin{array}{lll}\overrightarrow{ BA } & \overline{ n _{1}} & \overline{ n _{2}}\end{array}\right]=60+32+16=108$

$\left|\overline{n_{1}} \times \overline{n_{2}}\right|=\sqrt{100+64+16}=\sqrt{180}$

$S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}$

Any point on it $\vec{a}_{1}(-7,6,0)$

and $L_{1}$ is parallel to $\vec{b}_{1}(-6,7,1)$

$L_{2}: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}$

Any point on it, $\vec{a}_{2}(7,2,6)$

and $L_{2}$ is parallel to $b_{2}(-2,1,1)$

Shortest distance between $L_{1}$ and $L_{2}$

$=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|=\left|\frac{(-14,4,-6) \cdot(3,2,4)}{\sqrt{9+4+16}}\right|$

$=2 \sqrt{29}$

$3 l^{2}+ m ^{2}+ cl (1+ m )=0$

$n =1+ m$

$3 l ^{2}+ m ^{2}+ cl ^{2}+ clm =0$

$(3+ c ) l ^{2}+ clm + m ^{2}=0$

$(3+c)\left(\frac{l}{m}\right)^{2}+c\left(\frac{l}{m}\right)+1=0 \ldots \ldots(1)$

$\because$ lies are parallel.

Roots of $(1)$ must be equal

$\Rightarrow D=0$

$c ^{2}-4(3+ c )=0$

$c ^{2}-4 c -12=0$

$( c -6)( c +2)=0$

$c =6$ or $c =-2$

$+ve$ value of $c =6$

$(3 \lambda-2 \mu+7,-\lambda-3 \mu-6, \lambda-\mu-2)$

$\frac{3 \lambda-2 \mu+7}{1}=\frac{-\lambda-3 \mu-6}{-4}=\frac{\lambda-\mu-2}{2}$

Taking first ($2$) $-12 \lambda+8 \mu-28=-\lambda-3 \mu-6$

$\lambda-\mu+2=0$

Taking second and  third

$-2 \lambda-6 \mu-12=-4 \lambda+4 \mu+8$

$\lambda-5 \mu-10=0$

After solving above two equation $\lambda=-5, \mu=-3$

$A =(-8,6,7)$

$B =(-6,-2,-3)$

$\quad( AB )^{2}=4+64+16=84$

$\Rightarrow \frac{x-\frac{1}{8}}{1}=\frac{y-0}{-\sqrt{2}}=\frac{z-0}{0}=\lambda$

$-8 x-6 \sqrt{3} z=1, y=0$

$\Rightarrow \frac{x+\frac{1}{8}}{3 \sqrt{3}}=\frac{y-0}{0}=\frac{z-0}{-4}$

$\left|\begin{array}{lll}\frac{1}{4} & 0 & 0 \\ 1 & -\sqrt{2} & 0 \\ 3 \sqrt{3} & 0 & -4\end{array}\right|=\sqrt{2}$

$d=\frac{1}{\sqrt{51}}$

$\frac{1}{d^{2}}=51$

$\therefore M =(3 \lambda+6,2 \lambda+1,3 \lambda+2)$

Now, $\overrightarrow{ PM }=(3 \lambda+5) \hat{ i }+(2 \lambda-1) \hat{ j }+(3 \lambda-1) \hat{ k }$

$\because \overrightarrow{ PM } \perp(3 \hat{ i }+2 \hat{ j }+3 \hat{ k })$

$\therefore 3(3 \lambda+5)+2(2 \lambda-1)+3(3 \lambda-1)=0$

$\lambda=\frac{-5}{11}$

$\therefore M \left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)$

Since $R$ is mid-point of $PM$

$22(\alpha+\beta+\gamma)=125$

$\therefore$ equation of line is

$\overrightarrow{ r }=\hat{ i }+2 \hat{ j }+4 \hat{ k }+\lambda(\hat{ i }-4 \hat{ j }-3 \hat{ k })$

Let $A (1,2,4)$ and $P$ be $(1+\lambda, 2-4 \lambda, 4-3 \lambda)$

$\therefore \overline{ PA } \cdot(\hat{ i }-4 \hat{ j }-3 \hat{ k })=0$

$\lambda=\frac{1}{2}$3

$P \left(\frac{1}{2}, 2, \frac{-5}{2}\right)$

$| AP |=\sqrt{\frac{21}{2}}$

$(2 \lambda-5)^{2}+(3 \lambda-4)^{2}+(2 \lambda-6)^{2}=26$

$\lambda=1,3$

$Q (1,1,3) ; R (5,7,7) ; \quad P (4,2,7)$

Area of triangle $PQR =1 / 2|\overrightarrow{ PQ } \times \overrightarrow{ PR }|$

$=\sqrt{153}$

$4 \lambda-2-2 a+9 \lambda-3+\lambda+1=0$

$14 \lambda-4-2 a =0$

$7 \lambda-2-a=0$

and,

$(2 \lambda-1-a)^{2}+(3 \lambda-1)^{2}+(\lambda+1)^{2}=24$

$5 \lambda-1)^{2}+(3 \lambda-1)^{2}+(\lambda+1)^{2}=24$

$35 \lambda^{2}-14 \lambda-21=0$

$(\lambda-1)(35 \lambda+21)=0$

For, $\lambda=1 \quad \Rightarrow a =5$

Let $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ be reflection of point $P$

$\alpha_{1}+5=2 \quad \alpha_{2}+4=12 \quad \alpha_{3}+2=0$

$\alpha_{1}=-3 \quad \alpha_{2}=8 \quad \alpha_{3}=-2$

$a+\alpha_{1}+\alpha_{2}+\alpha_{3}=8$

$DR$ of $PQ \equiv\left(\frac{56}{17}+2, \frac{43}{17}+1, \frac{111}{17}-1\right)$

$\equiv\left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right)$

$\frac{90}{17} \alpha+\frac{60}{17}(-1)+\frac{94}{17} \beta=0$

$90 \alpha+94 \beta=60$

$\beta=\frac{60-90 \alpha}{94}$

$\beta=\frac{30(2-3 \alpha)}{94}$

$\beta=-30 \frac{(3 \alpha-2)}{94}$

$\beta=\frac{-15}{47}(3 \alpha-2)$

$\Rightarrow \frac{\beta}{-15}=\frac{3 \alpha-2}{47}$

$\Rightarrow \beta=-15, \alpha=-15$

$\alpha^{2}+\beta^{2}=225+225$

$=450$

$\Rightarrow P(x, y, z)=(2 r+1,3 r-1,-2 r+1)$

Since, $\overline{Q P} \perp(2 \hat{i}+3 \hat{j}-2 \hat{k})$

$\Rightarrow 4 r+2+9 r-6+4 r+2=0$

$\Rightarrow r =\frac{2}{17}$

$\Rightarrow P \left(\frac{21}{17}, \frac{-11}{17}, \frac{13}{17}\right)$

$\Rightarrow \overline{ PQ }=\frac{21 \hat{ i }-28 \hat{ j }-21 \hat{ k }}{17}$

So, $\overline{ QP }: \frac{ x }{-3}=\frac{ y -1}{4}=\frac{ z -2}{3}$

$3 y-2 z-1=0$ D.R's $\Rightarrow(0,3,-2)$

$3 x-z+4=0$ D. R's $\Rightarrow(3,-1,0)$

Let DR's of given line are $a, b, c$

Now $3 \mathrm{~b}-2 \mathrm{c}=0 \,\& 3 \mathrm{a}-\mathrm{c}=0$

$\therefore 6 \mathrm{a}=3 \mathrm{~b}=2 \mathrm{c}$

$a: b: c=3: 6: 9$

Any pt on line

$3 \mathrm{~K}-1,6 \mathrm{~K}+1,9 \mathrm{~K}+1$

Now $3(3 \mathrm{~K}-1)+6(6 \mathrm{~K}+1) 1+9(9 \mathrm{~K}+1)=0$

$\Rightarrow \mathrm{K}=\frac{1}{3}$

Point on line $\Rightarrow(0,3,4)$

Given point $(2,-1,6)$

$\Rightarrow \text { Distance }=\sqrt{4+16+4}=2 \sqrt{6}$

then shortest distance between two lines is

$L=\frac{(\bar{a}-\vec{c}) \cdot(\bar{b} \times \bar{d})}{|b \times d|}$

$\therefore \vec{a}-\vec{c}=((\alpha+4) \hat{i}+2 \hat{j}+3 k)$

$\frac{\vec{b} \times \bar{d}}{|b \times d|}=\frac{(2 \hat{i}+2 \hat{j}+k)}{3}$

$\therefore((\alpha+4) \hat{i}+2 \hat{j}+3 \hat{j}) \cdot \frac{(2 \hat{i}+2 \hat{j}+k)}{3}=9$

or $\alpha=6$

$\frac{x+2}{3}=y=\frac{z}{3 / b}$

lines are Co-planar

$\left|\begin{array}{ccc}a & 1 & a \\ 3 & 1 & \frac{3}{b} \\ -1 & 0 & -1\end{array}\right|=0 \Rightarrow-\left(\frac{3}{b}-a\right)-1(a-3)=0$

$a-\frac{3}{b}-a+3=0$

$b=1, a \in R-\{0\}$

$L_{2}: \frac{(x-2)}{1}=\frac{y-\lambda}{2}=\frac{z-3}{4} \quad \vec{r}_{2}=\hat{i}+2 \hat{j}+4 \hat{k}$

Shortest distance $=$ Projection of $\vec{a}$ on $\vec{r}_{1} \times \vec{r}_{2}$

$=\frac{\left|\vec{a} \cdot\left(\vec{r}_{1} \times \vec{r}_{2}\right)\right|}{\left|\vec{r}_{1} \times \vec{r}_{2}\right|}$

$\left|\overrightarrow{\mathrm{a}} \cdot\left(\vec{r}_{1} \times \vec{r}_{2}\right)\right|=\left|\begin{array}{ccc}1 & \lambda-2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{array}\right|=|14-5 \lambda|$

$\left|\vec{r}_{1} \times \vec{r}_{2}\right|=\sqrt{38}$

$\therefore \frac{1}{\sqrt{38}}=\frac{|14-5 \lambda|}{\sqrt{38}}$

$\Rightarrow 14-5 \lambda \mid=1$

$\Rightarrow 14-5 \lambda=1 \text { or } 14-5 \lambda=-1$

$\Rightarrow \lambda=\frac{13}{5} \text { or } 3$

$\therefore$ Integral value of $\lambda=3$

$\{\therefore$ Shortest distance between then is zero $\}$

$(k+1)[2-6]-4[1-9]+6[2-6]=0$

$\mathrm{~K}=1$

Now, $\ell^{2}+ m ^{2}= n ^{2}=(\ell+ m )^{2}$

$\Rightarrow 2 \ell m =0$

If $\ell=0 \Rightarrow m = n =\pm \frac{1}{\sqrt{2}}$

And, If $m=0 \Rightarrow n=\ell=\pm \frac{1}{\sqrt{2}}$

So, direction cosines of two lines are

$\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$

Thus, $\cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$

$\ell \mathrm{m}+\mathrm{n}(\ell+\mathrm{m})=0$

$\ell \mathrm{m}+2(\ell+\mathrm{m})^{2}=0$

$2 \ell^{2}+2 \mathrm{~m}^{2}+5 \mathrm{~m} \ell=0$

$2\left(\frac{\ell}{\mathrm{m}}\right)^{2}+2+5\left(\frac{\ell}{\mathrm{m}}\right)=0$

$2 \mathrm{t}^{2}+5 \mathrm{t}+2=0$

$(\mathrm{t}+2)(2 \mathrm{t}+1)=0$

$\Rightarrow \mathrm{t}=-2 ;-\frac{1}{2}$

$(i)$ $\frac{\ell}{\mathrm{m}}=-2$

$\frac{\mathrm{n}}{\mathrm{m}}=-2$

$(-2 \mathrm{~m}, \mathrm{~m},-2 \mathrm{~m})$

$(-2,1,-2)$

$(ii)$ $\frac{\ell}{m}=-\frac{1}{2}$

$n=-2 \ell$

$(\ell,-2 \ell,-2 \ell)$

$(1,-2,-2)$

$\cos \theta=\frac{-2-2+4}{\sqrt{9} \sqrt{9}}=0 \Rightarrow 0=\frac{\pi}{2}$

$Q (1,2,-4)$

$\overline{ PR } \| 4 \hat{ i }-\hat{ j }+2 \hat{ k }$

$\overline{ QS } \|-2 \hat{ i }+\hat{ j }-2 \hat{ k }$

dr's of normal to the plane containing

$P , T $ and $Q$ will be proportional to :

$\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 4 & -1 & 2 \\ -2 & 1 & -2\end{array}\right|$

$\therefore \quad \frac{\ell}{0}=\frac{ m }{4}=\frac{ n }{2}$

For point, $T: \overline{P T}=\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$

$\overline{ QT }=\frac{x-1}{-2}=\frac{y-1}{1}=\frac{z+4}{-2}=\mu$

$T :(4 \lambda+3,-\lambda-1,2 \lambda+2)$

$\quad \cong(2 \mu+1, \mu+2,-2 \mu-4)$

$4 \lambda+3=-2 \mu+1 \quad \Rightarrow 2 \lambda+\mu=-1$

$\lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$

$ \mu=-5 \quad \lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$

So point $T :(11,-3,6)$

$\overline{ OA }=(11 \hat{ i }-3 \hat{ j }+6 \hat{ k }) \pm\left(\frac{2 \hat{ j }+\hat{ k }}{\sqrt{5}}\right) \sqrt{5}$

$\overline{ OA }=(11 \hat{ i }-3 \hat{ j }+6 \hat{ k }) \pm(2 \hat{ j }+\hat{ k })$

$\overline{ OA }=11 \hat{ i }-\hat{ j }+7 \hat{ k }$

$9 \hat{ i }-5 \hat{ j }+5 \hat{ k }$

$|\overline{ OA }|=\sqrt{121+1+49}=\sqrt{171}$

$\sqrt{81+25+25}=\sqrt{131}$

$\frac{3-a}{\ell}=\frac{5-2}{3}=\frac{7-b}{4}$

$\frac{3- a }{\ell}=1$ And $\frac{7-b}{4}=1$

$a+\ell=3 \quad \ldots(1)$ And $b=3$

$\overrightarrow{ v }_{1}=<4,3,8>-<3,5,7>$

$\overrightarrow{ v }_{1}=<1,-2,1>$

$\overrightarrow{ v }_{2}=\langle\ell, 3,4\rangle$

$\overrightarrow{ v }_{1} \cdot \overrightarrow{ v }_{2}=0 \quad \Rightarrow \quad \ell-6+4=0 \quad \Rightarrow \quad \ell=2$

$a +\ell=3 \Rightarrow a =1$

$L _{1}: \frac{ x -1}{2}=\frac{ y -2}{3}=\frac{ z -3}{4}$

$L _{2}: \frac{ x -2}{3}=\frac{ y -4}{4}=\frac{ z -5}{5}$

$A =<1,2,3>$

$B =<2,4,5>$

$\overline{ AB }=<1,2,2>$

$\overrightarrow{ p }=2 \hat{ i }+3 \hat{ j }+4 \hat{ k }$

$\overrightarrow{ q }=3 \hat{ i }+4 \hat{ j }+5 \hat{ k }$

$\overrightarrow{ p } \times \overrightarrow{ q }=-\hat{ i }+2 \hat{ j }-\hat{ k }$

Shortest distance $=\left|\frac{\overline{ AB } \cdot(\overrightarrow{ p } \times \overrightarrow{ q })}{|\overrightarrow{ p } \times \overrightarrow{ q }|}\right|=\frac{1}{\sqrt{6}}$

$\frac{ x -3}{7}=\frac{ y -2}{5}=\frac{ z -1}{-9}$

$Q =(20, b ,- a -9)$

$\frac{\frac{20+a}{2}-3}{7}=\frac{\frac{b+6}{2}-2}{5}=\frac{-\frac{9}{2}-1}{-9}$

$\frac{14+9}{14}=\frac{b+2}{10}=\frac{a+2}{18}$

$\Rightarrow a=-56$ and $b=-32$

$\Rightarrow \quad|a+b|=88$

$\frac{x-0}{1}=\frac{y+2 \lambda}{1}=\frac{z-\lambda}{1}$

Shortest distance $=\frac{\left( a _{2}- a _{1}\right) \cdot\left( b _{1} \times b _{2}\right)}{\left| b _{1} \times b _{2}\right|}$

$b _{1} \times b _{2}=\left|\begin{array}{ccc} i & j & k \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 1\end{array}\right|$

$=\hat{ i }\left(\frac{1}{2}+\frac{1}{2}\right)-\hat{ j }\left(1+\frac{1}{2}\right)+\hat{ k }\left(1-\frac{1}{2}\right)$

$=\hat{ i }-\frac{3}{2} \hat{ j }+\frac{\hat{ k }}{2}=\frac{2 \hat{ i }-3 \hat{ j }+\hat{ k }}{2}$

$\frac{b_{1} \times b_{2}}{\left|b_{1} \times b_{2}\right|}=\frac{2 \hat{i}-3 \hat{j}+\hat{k}}{\sqrt{14}}$

$\frac{\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)}{\left|b_{1} \times b_{2}\right|}=\left(-\lambda \hat{i}+\left(-2 \lambda+\frac{1}{2}\right)+\lambda \hat{k}\right)$

$\quad\quad\quad\quad\quad\quad\quad\left(\frac{2 \hat{i}-3 \hat{j}+\hat{k}}{\sqrt{14}}\right)$

$=\left|\frac{-2 \lambda+6 \lambda-\frac{3}{2}+\lambda}{\sqrt{14}}\right|=\frac{\sqrt{7}}{2 \sqrt{2}}$

$\left|5 \lambda-\frac{3}{2}\right|=\frac{7}{2}$

$5 \lambda=\frac{3}{2} \pm \frac{7}{2}$

$5 \lambda=5,-2$

$\lambda=1,-\frac{2}{5}$

$\ell_{2}: \overrightarrow{ r }=(3+2 s ) \hat{ i }+(3+2 s ) \hat{ j }+(4+ s ) \hat{ k }$

$DR$ of $\ell_{1} \equiv(1,2,2)$

$DR$ of $\ell_{2} \equiv(2,2,1)$

$DR$ of $\ell$ (line $\perp$ to $\left.\ell_{1} \& \ell_{2}\right)$ $=(-2,3,-2)$

$\therefore \ell: \overrightarrow{ r }=-2 \mu \hat{ i }+3 \hat{ j }-2 \mu \hat{ k }$

for intersection of $\ell \& \ell_{1}$

$3+t=-2 \mu$

$-1+2 t=3 \mu$

$4+2 t=-2 \mu$

$\Rightarrow t =-1 \& \lambda=-1$

$\therefore$ Point of intersection $P \equiv(2,-3,2)$

Let point on $\ell_{2}$ be $Q(3+2 s, 3+2 s, 2+s)$

Given $PQ =\sqrt{17} \quad \Rightarrow( PQ )^{2}=17$

$\Rightarrow(2 s +1)^{2}+(6+2 s )^{2}+( s )^{2}=17$

$\Rightarrow 9 s ^{2}+28 s +20=0$

$\Rightarrow s =-2,-\frac{10}{9}$

$s \neq-2$ as point lies on $1^{\text {st }}$ octant.

$\therefore a =3+2\left(-\frac{10}{9}\right)=\frac{7}{9}$

$b =3+2\left(-\frac{10}{9}\right)=\frac{7}{9}$

$c =2+\left(-\frac{10}{9}\right)=\frac{8}{9}$

$\therefore 18(a+b+c)=18\left(\frac{22}{9}\right)=44$

$=\left|\frac{(\hat{i}-3 \hat{j}+8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+7 \hat{k})}{9}\right|=8$

$=\sqrt{270}=3 \sqrt{30}$

$(2 \lambda-1,-2 \lambda+3,-\lambda)$

Direction ratio of $PQ \equiv(2 \lambda-2,-2 \lambda+1,3-\lambda)$

$PQ$ is $\perp$ to line

$\Rightarrow 2(2 \lambda-2)-2(-2 \lambda+1)-1(3-\lambda)=0$

$4 \lambda-4+4 \lambda-2-3+\lambda=0$

$9 \lambda=9 \Rightarrow \lambda=1$

$\Rightarrow \quad$ Point $R$ is (1,1,-1)

$\begin{array}{l|l|l} \frac{a+1}{2}=1 & \frac{b+2}{2}=1 & \frac{c-3}{2}=-1 \\ a=1 & b=0 & c=1 \end{array}$

$\Rightarrow a+b+c=2$

D.R. of $\mathrm{AP}=<\frac{5}{3}-1, \frac{7}{3}-0, \frac{17}{3}-3>$

$\mathrm{BP} \perp \mathrm{AP}$

$\Rightarrow \alpha=4$

$\overrightarrow{ r }=\hat{ i }(2+ m )+\hat{j}( m -1)+\hat{ k }(- m )$

For intersection

$\begin{array}{ll}1+2 \ell=2+m & \ldots \ldots (i)\\ -1=m-1 & \ldots \ldots (ii)\\ \ell=-m & \ldots \ldots(iii)\end{array}$

from (ii) $m =0$

from (iii) $\ell=0$

These values of $m$ and $\ell$ do not satisfy equation ( 1 ).

Hence the two lines do not intersect for any

values of $\ell$ and $m$.

$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \gamma=1$

$\Rightarrow \cos ^{2} \gamma=1-\frac{3}{4}=\frac{1}{4}$

$\Rightarrow \cos \gamma=\pm \frac{1}{2}$

$\Rightarrow \gamma=\frac{\pi}{3} \text { or } \frac{2 \pi}{3}$

A point on this line is $A(0,1,-1)$

$\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=0$

We get $\lambda=\frac{-1}{2}$

$\therefore C=\left(-\frac{1}{2}, 1, \frac{-1}{2}\right)$

$|\overrightarrow{\mathrm{BC}}|=\sqrt{\frac{2}{3}}$

$\sqrt{\left(\beta+\frac{1}{2}\right)^{2}+\left(1^{2}+\left(\beta+\frac{1}{2}\right)\right)^{2}}=\sqrt{\frac{2}{3}}$

$\therefore \beta=0,-1$

$\beta=-1 \quad(\beta \neq 0)$

$\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$

Line $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$

$\Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=\frac{z}{1}$

Given both the lines are perpendicular

$\Rightarrow a a^{\prime}+c^{\prime}+c=0$

$\overrightarrow{\mathrm{AB}}=(3 \mu+2) \mathrm{i}+(3-\mu) \mathrm{j}+(4-5 \mu)$

${\hat k}$ which is parallel to plane $x-4 y+3 z=1$

$\Rightarrow 1(3 \mu+2)-4(3-\mu)+3(4-5 \mu)=0$

$=-8 \mu+2=0 \Rightarrow \mu=\frac{1}{4}$

$\Rightarrow \quad \lambda=\frac{1}{2}$

$\therefore $ Length of perpendicular

$(=P M)=\sqrt{0+\frac{1}{4}+\frac{49}{4}}$

$=\sqrt{\frac{50}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{\sqrt{2}}$

which is greater than 3 but less than 4

$\mathrm{R}(4, \mathrm{y}, \mathrm{z}) \text { lies on this }$

$\Rightarrow \frac{1}{3}=\frac{y+3}{3}=\frac{z-4}{6}$

$\Rightarrow \mathrm{R}(4,-2,6)$

$\mathrm{QR}=\sqrt{16+4+36}=2 \sqrt{14}$

Now $A D \perp B C$

$\mathrm{D} \mathrm{R}$   of     $BC$

$\Rightarrow a_{1}=3, b_{1}=0, c_{1}=y$

$D \cdot R,$ of $A D$

$\Rightarrow a_{2}=3 \lambda-3, b_{2}=2, c_{2}=4 \lambda-2$

$\Rightarrow a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

$\Rightarrow 3(3 \lambda-3)+0+4(4 \lambda-2)=0$

$25 \lambda=17$

$\Rightarrow \lambda=\frac{17}{25}$

$\mathrm{Co}-$ ordinate of point $\mathrm{D}$

$\left(\frac{1}{25}, 1, \frac{68}{25}\right)$

$\mathrm{AD}=\sqrt{\frac{576}{625}+4+\frac{324}{25}}=\frac{2}{5} \sqrt{34}$

Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}$

$=\frac{1}{2} \times 5 \times \frac{2}{6} \sqrt{34}$

$=\sqrt{34}$

$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ are $2,2,1$

Also second line can be written as:

$\frac{{\frac{{x - 5}}{2} = \frac{{y - 2}}{P} = \frac{{ - 3}}{4}}}{7}$

$\therefore $ its direction cosines are $2 . \frac{P}{7} .4$

$\text { Also, } \cos \theta=\frac{2}{3} \quad \text { (Given) }$

$\therefore \cos \theta  = | - \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}$

$ \Rightarrow \frac{2}{3} = \frac{{2 \times 2 + \left( {2 \times \begin{array}{*{20}{l}}
P\\
7
\end{array}} \right) + (1 \times 4)}}{{\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + \frac{{{P^2}}}{{49}}}  + {4^2}}}$

${=\frac{4+\frac{2 P}{7}+4}{3 \times \sqrt{2^{2}+\frac{P^{2}}{49}+4^{2}}}} $

${\Rightarrow\left(4+\frac{P}{7}\right)^{2}=20+\frac{P^{2}}{49}} $

${\Rightarrow 16+\frac{8 P}{7}+\frac{P^{2}}{49}=20+\frac{P^{2}}{49}} $

${\Rightarrow \frac{8 P}{7}-4 \Rightarrow P=\frac{7}{2}}$

Now, $A^{\prime} B^{\prime}=B C=\sqrt{A B^{2}-A C^{2}}=\sqrt{2-\frac{4}{3}}=\sqrt{\frac{2}{3}}$

Length of projection $=\sqrt{\frac{2}{3}}$

$l+3 m+5 n=0$      ....$(1)$

and $5 l m-2 m n+6 n l=0$      .....$(2)$

From eq. $( 1 )$ we have $l=-3 m-5 n$

Put the value of $l$ in eq. $(2),$ we get;

$5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0$

$\Rightarrow 15 m^{2}+45 m n+30 n^{2}=0$

$\Rightarrow m^{2}+3 m n+2 n^{2}=0$

$\Rightarrow m^{2}+2 m n+m n+2 n^{2}=0$

$\Rightarrow(m+n)(m+2 n)=0$

$\therefore m=-n$ or $m=-2 n$

For $m=-n, l=-2 n$

And for $m=-2 n, l=n$

$\therefore(l, m, n)=(-2 n,-n, n)$ Or $(l, m, n)$

$=(n,-2 n, n)$

$\Rightarrow(l, m, n)=(-2,-1,1)$ Or $(l, m, n)$

$=(1,-2,1)$

Therefore, angle between the lines is given as:

$\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}}$

$\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)$

$\left| {\begin{array}{*{20}{c}}
{3 - 1}&{2 - 2}&{1 - \left( { - 3} \right)}\\
1&2&{{\lambda ^2}}\\
1&{{\lambda ^2}}&2
\end{array}} \right| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&4\\
1&2&{{\lambda ^2}}\\
1&{{\lambda ^2}}&2
\end{array}} \right. = 0$

$\Rightarrow 2\left(4-\lambda^{4}\right)+4\left(\lambda^{2}-2\right)=0$

$\Rightarrow 4-\lambda^{4}+2 \lambda^{2}-4=0$

$\Rightarrow \lambda^{2}\left(\lambda^{2}-2\right)=0$

$\Rightarrow \lambda=0, \sqrt{2},-\sqrt{2}$

$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(2+4 \lambda) z-(3+4 \lambda)=0$

If this plane is parallel to z-axis then normal to the plane will be perpendicular to z-axis.

$\therefore$ $(1+2 \lambda)(0)+(1+3 \lambda)(0)+(2+4 \lambda)(1)=0$

$\lambda=-\frac{1}{2}$

Thus, Required plane is

$(x+y+2 z-3)-\frac{1}{2}(2 x+3 y+4 z-4)=0$

$\Rightarrow y+2=0$

$\therefore \mathrm{S} . \mathrm{D}=\frac{2}{\sqrt{(1)^{2}}}=2$

Eliminationg $n$ from both the equations, we have 

 ${l^2} + {m^2} - {\left( {l + m} \right)^2} = 0$

$ \Rightarrow {l^2} + {m^2} - {l^2} - {m^2} - 2ml = 0$

$ \Rightarrow 2lm = 0$

$ \Rightarrow lm = 0$

$ \Rightarrow l = 0\,\,\,\,or\,\,\,m = 0$

If $l=0$, we have $m+n=0$ and ${m^2} - {n^2} = 0$

$ \Rightarrow l = 0,m = \lambda ,n =  - \lambda $

If $m=0$, we have $l+m=0$ and ${l^2} - {m^2} = 0$

$ \Rightarrow l =  - \lambda ,m = 0,n = \lambda $

So, the vector parallel to these given lines

are $\vec a = \hat j - \hat k\,$ and $\,\,\vec b =  - \hat i + \hat k$

If angle between the lines is $'\theta ',$ then

$\cos \theta  = \frac{{\left| {\vec a.\vec b} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} = \frac{1}{{\sqrt 2 .\sqrt 2 }}$

$ \Rightarrow \cos \theta  = \frac{1}{2}$

$\therefore \theta  = \frac{\pi }{3}$

$l=\cos \theta, m=\cos \theta, n=\cos (\pi-2 \theta)$

we have $l^{2}+m^{2}+n^{2}=1$

$\Rightarrow \cos ^{2} \theta+\cos ^{2} \theta+\cos ^{2}(\pi-2 \theta)=1$

$\Rightarrow 2 \cos ^{2} \theta+(-\cos 2 \theta)^{2}=1$

$\Rightarrow 2 \cos ^{2} \theta-1+\cos ^{2} 2 \theta=0$

$\Rightarrow \cos 2 \theta-[1+\cos 2 \theta]=0$

$ \Rightarrow \cos 2\theta  = 0\,\,or\,\cos 2\theta  =  - 1$

$\Rightarrow 2 \theta=\pi / 2$ or $2 \theta=\pi$

$\Rightarrow \theta=\pi / 4$ or $\theta=\frac{\pi}{2}$

$\Rightarrow \theta=\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$