Maths M.C.Q (1 Marks)

 ${l^2} + {m^2} + {n^2} = 1 \Rightarrow 3{l^2} = 1 \Rightarrow l = \frac{1}{{\sqrt 3 }} = m = n$

Now, $r = |r|(li + mj + nk) = 6\left( {\frac{1}{{\sqrt 3 }}i + \frac{1}{{\sqrt 3 }}j + \frac{1}{{\sqrt 3 }}k} \right)$

Hence,$r = 2\sqrt 3 (i + j + k)$.

$\Rightarrow \,\,l + n = 0$

and $l\,( - 2) + m(1) + n( - 4) = 0\,\,\,\, $

$\Rightarrow \,\,\,2l - m + 4n = 0$

$\therefore \,\,\,l = - \frac{m}{2} = - \,n$.

Hence direction ratios are $(1,\,\, - 2,\,\, - 1)$.

==> ${\cos ^2}\alpha + {\sin ^2}\alpha + {\cos ^2}\gamma = 1$

==> ${\cos ^2}\gamma + 1 = 1\,\, $

$\Rightarrow \,\,\gamma = {90^o}$.

${\left[ {\sqrt {({x^2} + {y^2})} } \right]^2} + {\left[ {\sqrt {({y^2} + {z^2})} } \right]^2} + {\left[ {\sqrt {({z^2} + {x^2})} } \right]^2} = 36$

$ \Rightarrow \,\,{x^2} + {y^2} + {z^2} = 18$

Then distance from origin to the point $(x, y, z)$ is

$\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {18} = 3\sqrt 2 $.

Then co-ordinates of $P$ are $\left( {\frac{{5\lambda + 2}}{{\lambda + 1}},\,\frac{{\lambda + 2}}{{\lambda + 1}},\,\frac{{ - 2\lambda + 1}}{{\lambda + 1}}} \right)$.

It is given that the $x$ - coordinate of $P$ is $4$.

i.e.,$\frac{{5\lambda + 2}}{{\lambda + 1}} = 4$

==> $\lambda = 2$

So, $z$-coordinate of $P$ is $\frac{{ - 2\lambda + 1}}{{\lambda + 1}} = \frac{{ - 4 + 1}}{{2 + 1}} = - 1$.

Now, $l, m, n$ for $PQ$ are $\frac{1}{3},\frac{2}{3},\,\frac{{ - 2}}{3}$

$\therefore $ Projection of $RS$ on $PQ$ = $\sum l({x_2} - {x_1})$

= $\frac{2}{3} - \frac{4}{3} - \frac{2}{3} = \frac{{ - 4}}{3}$.

$ \Rightarrow \,\,a - 1 = \pm \,4$

$ \Rightarrow \,\,a = - 3,\,\,5$

$CD = \sqrt {{{(a - 1)}^2} + {{(a + 5)}^2} + {{(a - 1)}^2}} = 6$

$ \Rightarrow \,\,{a^2} - 2a - 15 = 0\,\, $

$\Rightarrow \,\,a = - 3,\,\,5$

But common solution of $(i)$ and $(ii)$ is $-3.$

and that of $AC$ is $\frac{{x + 1}}{4} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 4}}$ .....$(ii)$

Hence $\angle A = {\cos ^{ - 1}}\left( {\frac{{12 + 0 - 12}}{{\sqrt {9 + 9} \sqrt {16 + 4 + 16} }}} \right) = {90^o}$.

$\therefore \,$ If $l, m, n$ are direction cosines,

then $\frac{l}{a} = \frac{m}{1} = \frac{n}{c}$.

Hence direction ratios are $(a, 1, c).$

$y = - \,6 \pm \left( {\frac{3}{{\sqrt {17} }}} \right)\,(\sqrt {17} )$

and $z = 10 \pm \,\left( {\frac{{ - 2}}{{\sqrt {17} }}} \right)\,\,\,(\sqrt {17} )$.

Hence the required co-ordinates are $(1,\, - 3,\,\,8)$ or $(5,\,\, - 9,\,\,12)$.

$ \Rightarrow \,\,a = 1,\,\,b = 5,\,\,c = - 9$;

$\therefore \,\,\sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {107} $.

From point $P$ and $Q$ draw $PM$ and $QN$ perpendicular on the given plane and $QR \perp MP.$

$|MP|\, = \frac{{0 + 1 + 0 - 3}}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \frac{{ - 2}}{{\sqrt 3 }}$, $|NQ|\, = \frac{{ - 2}}{{\sqrt 3 }}$

$|PQ|\, = \sqrt {{{(0 - 0)}^2} + {{(0 - 1)}^2} + {{(1 - 0)}^2}} = \sqrt 2 $

$|RP| = |MP| - |MR|\, = \,|MP| - |NQ| = 0$

$|NM|\, = \,|QR|\, = \sqrt {P{Q^2} - R{P^2}} = \sqrt {{{(\sqrt 2 )}^2} - 0} $$ = \,\sqrt 2 .$

If $D$ divides $AB$ in ratio $\lambda :1$ then $D = \left( {\frac{{3\lambda + 4}}{{\lambda + 1}},\,\frac{{5\lambda + 7}}{{\lambda + 1}},\,\frac{{3\lambda + 1}}{{\lambda + 1}}} \right).....(i)$

$D.r’s$ of $PD$ are $2\lambda + 3,\,5\lambda + 7,\, - 2$

$D.r’s$ of $AB$ are $ - 1,\, - 2,\,2$

$PD\, \bot \,AB$; $ - (2\lambda + 3) - 2(5\lambda + 7) - 4 = 0$

==> $\lambda = \frac{{ - 7}}{4}$

Putting the value of $\lambda $ in $(i)$,

we get the point $D\left( {\frac{5}{3},\,\frac{7}{3},\,\frac{{17}}{3}} \right)$.

$r = (i - j + 2k) + \lambda (3i + j + k)$

The position vector of any point $P $ which is a variable point on the line, is $(i - j + 2k) + \lambda (3i + j + k)$

 $\therefore \overrightarrow {AP} = \lambda (3i + j + k) \Rightarrow |\overrightarrow {AP} | = \lambda \sqrt {11} $

Now, if $\lambda $ $\sqrt {11} = 3\sqrt {11} $ i.e., $\lambda = 3$ then the position vector of the point $P$  is $10i + 2j + 5k$.

If $\lambda \sqrt {11} = - 3\sqrt {11} ,$ i.e., $\lambda = - 3$ then the position vector of the point $P$ is $ - 8i - 4j - k$.

$r = (i + 2j - k) + \lambda (2i + j + k)$

and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$,

therefore it is parallel to the vector

$b = (2i + j + k) \times (i + 2j + 3k)$= $(i - 5j + 3k)$

Hence, the equation of the required line is

$r = (i + 3j + 2k) + \lambda '(i - 5j + 3k)$

==> $r = (i + 3j + 2k) + \lambda ( - i + 5j - 3k)$,

where $\lambda = - \lambda '$.

Therefore, if ${p_1}$ and ${p_2}$ are the lengths of the perpendicular from the origin to the planes $r.(i + 2j - 2k) + 5 = 0$ and $r.(i + 2j - 2k) - 8 = 0$ respectively,

then the required distance is given by ${p_1} + {p_2} = \frac{5}{3} + \frac{8}{3} = \frac{{13}}{3}$unit.

where ${a_1} = 3i - 2j - 2k,\,\,\,{b_1} = i$

${a_2} = i - j + 2k,\,\,\,\,\,{b_2} = j$

$|{b_1} \times {b_2}|\, = \,|i \times j|\, = \,|k| = 1$

Now, $[({a_2} - {a_1})\;{b_1}\;{b_2}] = ({a_2} - {a_1}).({b_1} \times {b_2})$

$ = ( - 2i + j + 4k)(k) = 4$

 Shortest distance $ = \frac{{[({a_2} - {a_1})({b_1} - {b_2})]}}{{|{b_1} \times {b_2}|}} = \frac{4}{1} = 4$.

Where ${a_1} = 4i - 3j - k;\,\,\,\,{b_1} = i - 4j + 7k$

${a_2} = i - j - 10k;\,\,\,\,{b_2} = 2i - 3j + 8k$

$|{b_1} \times {b_2}| = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 4}&7\\2&{ - 3}&8\end{array}} \right| = - 11i + 6j + 5k$

Now $[({a_2} - {a_1})\,\,{b_1}\,\,{b_2}] = ({a_2} - {a_1}).({b_1} \times {b_2})$

$ = ( - 3i + 2j - 9k)( - 11i + 6j + 5k) = 0$

Therefore, shortest distance $ = \frac{{[({a_2} - {a_1})\,\,{b_1}\,\,{b_2}]}}{{|{b_1} \times {b_2}|}} = 0$.

$\therefore $ $3a + 2b + c = 0$ and $a + b - 2c = 0$

$\frac{a}{{ - 4 - 1}} = \frac{b}{{1 + 6}} = \frac{c}{{3 - 2}}$ or $\frac{a}{{ - 5}} = \frac{b}{7} = \frac{c}{1}$

In order to find a point on the required line we put $z = 0$ in the two given equation to obtain, $3x + 2y = 5$ and $x + y = 3$.

Solving these two equations, we obtain $x = - 1,\,y = 4$.

$\therefore $ Co-ordinates of point on required line are $( - 1,\,4,\,0)$.

Hence required line is$\frac{{x + 1}}{{ - 5}} = \frac{{y - 4}}{7} = \frac{{z - 0}}{1}$.

$\, \Rightarrow \,\,( - \,3 - 2,\,\,1 - 1,\,\,7 - ( - 3))\,\, $

$\Rightarrow \,\,( - \,5,\,\,0,\,\,10)$

Direction ratio of the line parallel to line $\frac{{x - 1}}{3} = \frac{y}{4} = \frac{{z + 3}}{5}$ are $({a_2},\,{b_2},\,\,{c_2})\,\, $

$\Rightarrow \,\,(3,\,\,4,\,\,5)$

Angle between two lines,

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

$\cos \theta = \frac{{( - \,5 \times 3) + (0 \times 4) + (10 \times 5)}}{{\sqrt {25 + 0 + 100} \sqrt {9 + 16 + 25} }}$

$\cos \theta = \frac{{35}}{{25\sqrt {10} }}\,\, $

$\Rightarrow \,\,\theta = {\cos ^{ - 1}}\left( {\frac{7}{{5\sqrt {10} }}} \right)$.

$\frac{{x - 5}}{3} = \frac{{y - 7}}{{ - 1}} = \frac{{z + 2}}{1} = {r_1}$, (say)

and $\frac{{x + 3}}{{ - 36}} = \frac{{y - 3}}{2} = \frac{{z - 6}}{4} = {r_2}$, (say)

$\therefore $$x = 3{r_1} + 5 = - 36{r_2} - 3$, $y = - {r_1} + 7 = 3 + 2{r_2}$

and $z = {r_1} - 2 = 4{r_2} + 6$

On solving, we get $x = 21,\,y = \frac{5}{3},\,z = \frac{{10}}{3}$.

Trick: Check through options.

$ = \frac{3}{7}\sqrt {35} $.

$\frac{{x - 1}}{l} = \frac{{y - 2}}{m} = \frac{{z + 4}}{n}$

Now, according to question, $3l - 16m + 7n = 0$ and $3l + 8m - 5n = 0$

Hence required line is, $\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z + 4}}{6}$.

Hence the lines are perpendicular.

$\Rightarrow 60-30 p-14 a=0$          ......$(1)$

Lines are perpendicular

$\therefore 4-12+2 p=0$          .......$(2)$

from $(1)$ and $(2)$

$P=4, a=\frac{-30}{7}$

$(2 \lambda+1,3 \lambda-1,4 \lambda+1) ; \quad \lambda \in \mathrm{R}$

Any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$ is

$(\mu+3,2 \mu+\mathrm{k}, \mu) ; \mu \in \mathrm{R}$

The given lines intersect if and only if the system of equations (in $\lambda $ and $ \mu$)

${2 \lambda+1=\mu+3} $         ........$(i)$

${3 \lambda-1=2 \mu+k} $           ........$(ii)$

${4 \lambda+1=\mu}$           ..........$(iii)$

has a unique solution.

Solving $(i)$ and $(iii),$ we get $\lambda=\frac{-3}{2}, \mu=-5$

From $(ii),$ we get $\frac{-9}{2}-1=-10+k \Rightarrow k=\frac{9}{2}$

$\overrightarrow {\rm{b}} :\widehat {\rm{i}} + 2\widehat {\rm{j}} + 3\widehat {\rm{k}};\quad \overrightarrow q  = 2\widehat {\rm{i}} + 3\widehat {\rm{j}} + 4\widehat {\rm{k}}$

${\rm{SD}} = \left| {\frac{{\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow p  \times \overrightarrow q } \right)}}{{\left| {\overrightarrow p  \times \overrightarrow q } \right|}}} \right|$

$=\left|\frac{-(\hat{i}+2 \hat{j}+2 \hat{k}) \times(\hat{i}-2 \hat{j}+\hat{k})}{|\hat{i}-2 \hat{j}+\hat{k}|}\right|=\frac{1}{\sqrt{6}}$

$ \Rightarrow \overrightarrow n  = \frac{{ - 6\widehat {\rm{i}} - 15\widehat {\rm{j}} + 3\widehat {\rm{k}}}}{{3\sqrt {30} }}$

$A = (3,8,3);B = ( - 3, - 7,6) \Rightarrow \overrightarrow {AB}  =  - 6\hat i - 15\hat j + 3\hat k$

$\mathrm{SD}=|\overrightarrow{\mathrm{AB}} \hat{\mathrm{n}}|=\frac{1}{3 \sqrt{30}}(36+225+9)=\frac{270}{\sqrt{270}}$

$=3 \sqrt{30}$

$\overrightarrow{\mathrm{r}}=(-1,0,-1)+\mathrm{k}(2,1,1)$

${\rm{S}}.{\rm{D}} = \left| {\frac{{(\hat i + \hat j + \hat k) \cdot (\hat i - 2\hat j + \hat k) \times (2\hat i + \hat j + \hat k)}}{{|(\hat i - 2\hat j + \hat k) \times (2\hat i + \hat j + \hat k)|}}} \right| = \frac{3}{{\sqrt {35} }}$

$\mathrm{L}_{1}: \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$

$\mathrm{L}_{2}: \overline{\mathrm{r}}=\overline{\mathrm{c}}+\mathrm{m} \overline{\mathrm{d}}$ is

$D = \frac{{\left| {(\bar a - \bar c) \cdot \left| {\bar b \times \bar d} \right|} \right|}}{{\left| {\bar b \times \bar d} \right|}} = \frac{1}{{\sqrt {14} }}$

$\mathrm{L}_{1} \equiv(3 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$

$\mathrm{L}_{2} \equiv(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+9 \hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})$

$\mathrm{SD}=| \frac{\left(\mathrm{a}_{2}-\mathrm{a}_{1}\right) \cdot\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right)}{\left|\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right|}$

Where $a_{1}=3 \hat{i}-15 \hat{j}+9 \hat{k} $ and $ a_{2}=-\hat{i}+\hat{j}+9 \hat{k}$

and $\quad \mathrm{b}_{1}=2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} $ and $ \overrightarrow{\mathrm{b}}_{2}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})$

$\overrightarrow{\mathrm{PQ}} \cdot(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=0$

$\overrightarrow {{\rm{PQ}}}  = (3\lambda  + 3\mu  + 6)\widehat {\rm{i}} + (16 - \lambda  - 2\mu )\widehat {\rm{j}} + (2 + \lambda  - 4\mu )\widehat {\rm{k}}$

$\Rightarrow 7 \mu+11 \lambda=-4 \quad $ and $ \quad 29 \mu+7 \lambda=22$

$\Rightarrow \lambda=-1 $ and $ \mu=1$

$\mathrm{P} $ and $ \mathrm{Q}$ are $(3,8,3) $ and $(-3,-7,6)$ respectively

Hence equation of $\mathrm{PQ}$ is $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-8}{5}=\frac{\mathrm{z}-3}{-1}$

$\Rightarrow 2\left(-\frac{1}{2}-6 \lambda\right)-2\left(\frac{3}{2}+4\right)+\left(\lambda-\frac{7}{2}\right)(9 \lambda-2)=0$

$\Rightarrow 18 \lambda^{2}-91 \lambda-16=0$

$\Rightarrow \lambda_{1}+\lambda_{2}=\frac{91}{18}=\frac{182}{36}$

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and

$\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ is given by,

$\frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}$

$\therefore \quad$ The shortest distance between given lines

are

$\frac{{\left| {\begin{array}{*{20}{c}} { - 2}&4&5\\ 2&2&1\\ { - 1}&8&4 \end{array}} \right|}}{{\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}$

$ = \left| {\frac{{0 - 36 + 90}}{{\sqrt {405} }}} \right| = \frac{{54}}{{20.1}} = 2.68$

$ \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} $

$ \Rightarrow \lambda+2=4 \mathrm{k}-3 $

$ -\lambda=3 \mathrm{k}-2 $

$ \Rightarrow \mathrm{k}=1, \lambda=-1 $

$ 8 \lambda+7=\mathrm{k}-2 $

$ \therefore \mathrm{P}=(1,1,-1)$

Projection of $2 \hat{i}-2 \hat{k}$ on $2 \hat{i}+3 \hat{j}+\hat{k}$ is

$ =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} $

$ \therefore l^2=8-\frac{4}{14}=\frac{108}{14} $

$ \Rightarrow 14 l^2=108$

$\mathrm{G}(1,2,2)$

Point of intersection $A$ of given lines is $(2,-6,0)$

$\mathrm{AG}=\sqrt{69}$

$|\mathrm{~b}||\mathrm{c}| \cos \alpha=-3|\mathrm{~b}|^2 $

$|\mathrm{c}| \cos \alpha=-12 \text {, as }|\mathrm{b}|=4 $

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2 $

$\cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} $

$|\mathrm{c}|^2=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^2$

$=64 \times \frac{3}{4}+144=192$

$|\mathrm{c}|^2 \cos ^2 \alpha=144 $

$192 \cos ^2 \alpha=144 $

$192 \sin ^2 \alpha=48$

$\therefore \overrightarrow{\mathrm{PR}} \cdot(2,3,4)=0 $

$(\alpha-2, \beta-3, \gamma-5) \cdot(2,3,4)=0 $

$\Rightarrow 2 \alpha+3 \beta+4 \gamma=4+9+20=33$

$\frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 $

$\left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) $

$=(22,0,7)=(a, b, c) $

$\therefore \sqrt{324+144+16}=22$

$ \frac{\mathrm{x}-1}{4}=\frac{\mathrm{y}+2}{-2}=\frac{\mathrm{z}-3}{4} $

$ \mathrm{pt}(4 \mathrm{t}+1,-2 \mathrm{t}-2,4 \mathrm{t}+3) $

$ \text { distance }^2=16 \mathrm{t}^2+4 \mathrm{t}^2+16 \mathrm{t}^2=81 $

$ \mathrm{t}= \pm \frac{3}{2} $

$ \operatorname{pt}(7,-5,9) $

$ \alpha^2+\beta^2+\gamma^2=155$

option ($1$)

$ \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda $

$ \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) $

$ \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} $

$ \mathrm{P}(2 \mu+3,3 \mu+2,2 \mu+3) $

$ \lambda+2=2 \mu+3 \quad 3 \mu+2=5 \lambda+4 $

$ \lambda=2 \mu+1 \quad 3 \mu=5 \lambda+2 $

$ 3 \mu=5(2 \mu+1)+2 $

$ 3 \mu=10 \mu+7 $

$ \mu=-1 \quad \lambda=-1$

Both satisfies ($P$)

$ \mathrm{P}(1,-1,1) $

$ \mathrm{L}_3 \equiv \frac{\mathrm{x}}{1 / 4}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{1} $

$ \mathrm{~L}_3=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{4}=\mathrm{k}$

Coordinates of $\mathrm{Q}(\mathrm{k}, 2 \mathrm{k}, 4 \mathrm{k})$

$ \text { DR's of } \mathrm{PQ}=<\mathrm{k}-1,2 \mathrm{k}+1,4 \mathrm{k}-1> $

$ \mathrm{PQ} \perp \text { to } \mathrm{L}_3 $

$ (\mathrm{k}-1)+2(2 \mathrm{k}+1)+4(4 \mathrm{k}-1)=0 $

$ \mathrm{k}-1+4 \mathrm{k}+2+16 \mathrm{k}-4=0 $

$ \mathrm{k}=\frac{1}{7} $

$ \mathrm{Q}\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right)$

$ \mathrm{Q}\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right) $

$ \mathrm{PQ}=\sqrt{\left(1-\frac{1}{7}\right)^2+\left(-1-\frac{2}{7}\right)^2+\left(1-\frac{4}{7}\right)^2} $

$ =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7} $

$ \mathrm{PQ}=\frac{3 \sqrt{14}}{7}$

Option $-3$ will satisfy

$ x=3 \lambda-6, y=2 \lambda, z=\lambda-1 $

$ \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu $          ........................($2$)

$ x=4 \mu+7, y=3 \mu+9, z=2 \mu+4 $

$ 3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13 $      ............$(3)\times 2$

$ 2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9 $          ............$(4)\times 3$

$ \quad 6 \lambda-8 \mu=26 $

$ \quad 6 \lambda-9 \mu=27 $

$ \quad-\quad+\quad- $

$ \quad \mu=-1 $

$ \Rightarrow 3 \lambda-4(-1)=13 $

$ \quad 3 \lambda=9 $

$ \quad \lambda=3 $

$ \text { int. point }(3,6,2) ;(7,8,9) $

$ d^2=16+4+49=69 $

Ans. $d^2+6=69+6=75$