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Maths 5 Marks Questions

Introduction To 3D Coordinate Geometry · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 18 questions.

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Question 15 Marks
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Answer
Here, A(a, b, c), B(b, c, a), C(c, a, b)
$\text{AB}=\sqrt{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2 + (\text{c}-\text{a})^2}$
$=\sqrt{\text{a}^2 +\text{b}^2 -2\text{ab} + \text{b}^2 +\text{c}^2 - 2\text{bc} +\text{c}^2 +\text{a}^2 -2\text{ac}}$
$\text{AB}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab} - 2\text{bc}- 2\text{ac}}$
$\text{BC}=\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2 + (\text{a}-\text{b})^2}$
$=\sqrt{\text{b}^2 +\text{c}^2 - 2\text{bc}+ \text{c}2 + \text{a}^2 -2\text{ca} +\text{a}^2 +\text{b}^2 -2\text{ab}}$
$\text{BC}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$
$\text{CA}=\sqrt{(\text{a}-\text{c})^2+(\text{b}-\text{a})^2 + (\text{c}-\text{b})^2}$
$=\sqrt{\text{a}^2 +\text{c}^2 - 2\text{ac}+ \text{b}2 + \text{a}^2 -2\text{ab} +\text{b}^2 +\text{c}^2 -2\text{bc}}$
$\text{CA}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$
Since, AB = BC = CA, so
$\triangle\text{ABC}$ is an isosceles A.
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Question 25 Marks
Show that the points A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.
Answer
Here,
$\text{AB}=\sqrt{(1+1)^2+(3-6)^2+(4-10)^2}$
$=\sqrt{4 +9+ 36}$
$=7\text{ units}$
$\text{BC}=\sqrt{(-1+7)^2+(6-4)^2+(0-7)^2}$
$=\sqrt{36+4+9}$
$=7\text{ units}$
$\text{CD}=\sqrt{(-7+5)^2+(4-1)^2+(7-1)^2}$
$=\sqrt{4 +9+ 36}$
$=7\text{ units}$
$\text{DA}=\sqrt{(-5-1)^2+(1-3)^2+(1-4)^2}$
$=\sqrt{36+4+9}$
$=\sqrt{52}$
$=7\text{ units}$
Since, AB = BC = CD = DA
So, ABCD is a rhombus.
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Question 35 Marks
Show that the points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
Answer
Let A = (0, 7, 10), B = (-1, 6, 6) and C = (-4, 9, 6)

$\text{AB}=\sqrt{(0+1)^2+(7-6)^2+(10-6)^2}$

$=\sqrt{(1)^2+(1)^2+(4)^2}$

$=\sqrt{18}$

$=3\sqrt{2}\text{ units}$

$\text{BC}=\sqrt{(-1+4)^2+(6-9)^2+(6-6)^2}$

$=\sqrt{(3)^2+(3)^2+(0)}$

$=\sqrt{18}$

$=3\sqrt{2}\text{ units}$

$\text{AC}=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}$

$=\sqrt{(4)^2+(-2)^2+(4)^2}$

$=\sqrt{36}$

$=6\text{ units}$

$(\text{AB})^2+\text{(BC)}^2$

$=(3\sqrt{2})^2+(3\sqrt{2})^2$

$=18+18$

$=36$

$=\text{(AC)}^2$

Also (AB) = (BC)

Hence (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
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Question 45 Marks
Using distance formula prove that the following points are collinear:
P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
Answer
P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

$\text{PQ}=\sqrt{(0-1)^2+(7-4)^2+(-7+5)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

$\text{QR}=\sqrt{(1+1)^2+(4-10)^2+(-5+9)^2}$

$=\sqrt{(2)^2+(-6)^2+(4)^2}$

$=\sqrt{4+36+16}$

$=2\sqrt{14}\text{ units}$

$\text{PR}=\sqrt{(0+1)^2+(7-10)^2+(-7+9)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

Since, PQ + PR = QR

so, P, Q, R are collinear.
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Question 55 Marks
Find the coordinates of the point which is equidistant from the four points $O(0, 0, 0), A(2, 0, 0), B(0, 3, 0)$ and $C(0, 0, 8).$
Answer
Let the required point be $P(x_1y_1z)$
Here, $0(0, 0, 0), A(2, 0, 0), B(0, 3, 0), C(0, 0, 8)$
Since,$ (OP)^2 = (PA)^2$
$(x - 0)^2 + (Y - 0)^2 + (z - 0)^2 = (x - 2)^2 + (y - 0)^2 + (z- 0)^2$
$x^2 + y^2 + z^2 = x^2 - 4x + 4 + y^2 + z^2$
$4x = 4$
$X = 1$
$(OP)^2 = (PB)^2$
$(x - 0)^2 + (y - 0)^2 + (z - 0)^2 = (x - 0)^2 + (y - 3)^2 + (z - 0)^2$
$x^2 + y^2 + z^2 = x^2 + y^2- 6y + 9 + z^2$
$6y = 9$
$\text{y}=\frac{3}{2}$
$(OP)^2 = (PC)^2$
$(x - 0)^2 + (y - 0)^2 + (z - 0)^2 = (x - 0)^2 + (Y - 0)^2 + (z - 8)^2$
$x^2+ y^2+ z^2= x^2+ y^2+ z^2 - 16z + 64$
$16z = 64$
$z = 4$
The required point $=\Big(1,\ \frac{3}{2},\ 4\Big)$
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Question 65 Marks
Show that the points (3, 2, 2), (-1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.
Answer
Here,
$\text{OA}=\sqrt{(1-3)^2+(3-2)^2+(4-2)^2}$
$=\sqrt{4+1+4}$
$=3\text{ units}$
$\text{OB}=\sqrt{(1+1)^2+(3-1)^2+(4-3)^2}$
$=\sqrt{4+4+1}$
$=3\text{ units}$
$\text{OC}=\sqrt{(1-0)^2+(3-5)^2+(4-6)^2}$
$=\sqrt{1+4+4}$
$=3\text{ units}$
$\text{OD}=\sqrt{(1-2)^2+(3-1)^2+(4-2)^2}$
$=\sqrt{1+4+4}$
$=3\text{ units}$
Since, OA = OC = OD = OB, points A, B, C, O lie on a sphere with centre O.
Radius = 3 units.
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Question 75 Marks
Prove that the tetrahedron with vertices at the points O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0) is a regular one.
Answer
Here,
$\text{AB}=\sqrt{(0-1)^2+(1-0)^2+(1-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{BC}=\sqrt{(1-1)^2+(0-1)^2+(1-0)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{CA}=\sqrt{(1-0)^2+(1-1)^2+(0-1)^2}$
$=\sqrt{1+0+1}$
$=\sqrt{2}\text{ units}$
$\text{DA}=\sqrt{(0-0)^2+(0-1)^2+(0-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{OB}=\sqrt{(0-1)^2+(0-0)^2+(0-1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
$\text{DA}=\sqrt{(0-1)^2+(0-1)^2+(0-0)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}\text{ units}$
Since, OA = OB = OC = AB = BC = CA
So, O, A, B, C represent a regular tetrahedron.
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Question 85 Marks
Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.
Answer
Here

$\text{AB}=\sqrt{(1+5)^2+(3-5)^2+(0-2)^2}$

$=\sqrt{36+4+4}$

$=\sqrt{44}$

$=2\sqrt{11}\text{ units}$

$\text{BC}=\sqrt{(-5+9)^2+(5+1)^2+(2-2)^2}$

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{CD}=\sqrt{(-9+3)^2+(-1+3)^2+(2-0)^2}$

$=\sqrt{36+4+4}$

$=2\sqrt{11}\text{ units}$

$\text{DA}=\sqrt{(-3-4)^2+(-3-3)^2+0}$

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{AC}=\sqrt{(1+9)^2+(3+1)^2+(0-2)^2}$

$=\sqrt{150+16+4}$

$=\sqrt{120}$

$=4\sqrt{5}\text{ units}$

$\text{BD}=\sqrt{(-3+5)^2+(-3-5)^2+(0-2)^2}$

$=\sqrt{4+64+4} $

$=\sqrt{72}$

$=6\sqrt{2}\text{ units}$

Since,

AB = CD and BC = DA

⇒ ABCD is a parallelgram = BD

but, $\text{AC}\neq\text{BD}$

⇒ ABCD is not a rectangles.
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Question 95 Marks
If $A (-2,2,3)$ and $B (13,-3,13)$ are two points. Find the locus of a point P which moves in such a way that $3 PA =2 PB$.
Answer
Let P be $\left( x _1 y _1 z \right)$,
Here, $A (-2,2,3), B (13,-3,13)$ and $3 PA =2 PB$
$\Rightarrow 3 \sqrt{(x+2)^2(y-2)^2+(z-3)^2}$
$=2 \sqrt{(x-13)^2+(y+3)^2+(z-13)^2}$
squaring both the sides,
$\Rightarrow 9\left[x^2+4 x+4+y^2+4-4 y+z^2+9-6 z\right]$
$=4\left[x^2+169-26 x+y^2+9+6 y+z^2+169-26 z\right]$
$\Rightarrow 9 x^2-4 x^2+36 x+104 x+36-676+9 y^2-4 y^2+36-36-36 y-24 y+9 z^2-4 z^2+81-676-54 z+6 y z=0$
$\Rightarrow 5 x^2+5 y^2+5 z 2+140 x-60 y+50 z-1235=0$
$\Rightarrow 5\left(x^2+y^2+z^2\right)+140 x-60 y+50 z-1235=0$
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Question 105 Marks
Using distance formula prove that the following points are collinear:
A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
Answer
A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

$\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

$=\sqrt{(4-5)^2+(-3+7)^2+(-1-6)^2}$

$=\sqrt{(-1)^2+(4)^2+(-7)^2}$

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

$\text{BC}=\sqrt{(5-3)^2+(-7-1)^2+(6+8)^2}$

$=\sqrt{(2)^2+(-8)^2+(14)^2}$

$=\sqrt{4+64+196}$

$=\sqrt{264}$

$=2\sqrt{66}\text{ units}$

$\text{AC}=\sqrt{(4-3)^2+(-3-1)^2+(-1+8)^2}$

$=\sqrt{(1)^2+(-4)^2+(7)^2}$

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

Since, AC + AB = BC

so, A, B, C are collinear.
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Question 115 Marks
If $A(-2, 2, 3)$ and $B(13, -3, 13)$ are two points. Find the locus of a point P which moves in such a way that $3PA = 2PB.$
Answer
Let P be $(x_1 y_1 z),$
Here, $A(-2, 2, 3), B(13, -3, 13)$ and $3PA = 2PB$
$\Rightarrow3\sqrt{(\text{x}+2)^2(\text{y}-2)^2+(\text{z}-3)^2}$
$=2\sqrt{(\text{x}-13)^2+(\text{y}+3)^2+(\text{z}-13)^2}$
squaring both the sides,
$\Rightarrow 9[x^2+ 4x + 4 + y^2+ 4 - 4Y + z^2+ 9 - 6z]$
$= 4[x^2 + 169 - 26x + y^2 + 9 + 6y + z^2 + 169 - 26z]$
$\Rightarrow 9x^2 - 4x^2+ 36x + 104x + 36 - 676 + 9y^2 - 4y^2 + 36 - 36 -36y - 24y + 9z^2- 4z^2 + 81 - 676 - 54z + 6yz = 0$
$\Rightarrow 5x^2 + 5y^2 + 5z2 + 140x - 60y + 50z - 1235 = 0$
$\Rightarrow 5(x^2 + y^2 + z^2) + 140x - 60y + 50z - 1235 = 0$
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Question 125 Marks
Find the locus of $P$ if $PA^2 + PB^2 = 2k^2$, where A and B are the points $(3, 4, 5)$ and $(-1, 3, -7).$
Answer
Let $P(x_1, y_1, z),$
Here, $A(3, 4, 5), B(-1, 3, -7)$
$PA^2 + PB^2 = 2k^2$
$\Rightarrow (x - 3)^2 + (y - 4)^2 + (z - 5)^2 + (x + 1)^2 + (y - 3)^2 + (2 + 7)^2 = 2k^2$
$\Rightarrow x^2 + 9 - 6x + y^2 + 16 - 8y + z^2 + 25 - 10z + x^2 + 1 + 2x + y^2 + 9 - 6y + z^2+ 49 + 14z = 2k^2$
$\Rightarrow 2x^2 + 2y^2 + 2z^2- 4x - 14y + 4z + 109 = 2k^2$
$\Rightarrow 2(x^2 + y^2 + z^2) - 4x -14y + 4z + 109 - 2k^2 = 0$
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Question 135 Marks
Show that the plane $ax+ by+ cz + d = 0$ divides the line joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+d}{\text{ax}_2+\text{by}_2+\text{cz}_2+d}.$
Answer
Assume ratio is $\lambda:1$
Plane is $ax + by + cz + d = 0$
points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$
Assume point of intersection of line and plane is $D$
$\text{D}=\Big(\frac{\lambda\text{x}_2+\text{x}_1}{\lambda+1},\frac{\lambda\text{y}_2+\text{y}_1}{\lambda+1},\frac{\lambda\text{z}_2+\text{z}_1}{\lambda+1}\Big)$
As D lies on plane, substitute D in plane equation, we get
$\lambda(\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d})+\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}=0$
$\Rightarrow\lambda=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
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Question 145 Marks
Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelopiped so formed.
Answer
IMAGE

Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

$\therefore$ PA = Distance between planes PBEC and QDAF

= 5 - 3

= 2

PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.

$\therefore$ PB = 0 - (-2)

= 2

PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.

$\therefore$ PC = 2 - 5

= -3
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Question 155 Marks
Determine the points yz-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
Answer
Let $Q(0, y, z)$ be the required point.
So
$(AQ)^2 = (BQ)^2$
$\Rightarrow (0 - 1)^2 + (y + 1)^2 - (z - 0)^2 = (0 - 2) + (y + 1)^2 + (z - 2)^2$
$\Rightarrow 1 + y^2 + 1 + 2y + z^2 = 4 + y^2 + 1 - 2y + z^2 + 4 - 42$
$\Rightarrow 4y + 4z = 7 ... (i)$
$(BQ)^2 = (CQ)^2$
$\Rightarrow (0 - z)^2+ (y - 1)^2 + (z - 2)^2 = (0 - 3)^2 + (y - 2)^2 (2 + 1)^2$
$\Rightarrow 4 + y^2+ 1 - 2y + z^2+ 4 - 4z - 9 + y^2+ 4 - 4y + z^2+ 1 + 2z$
$\Rightarrow 2y - 6z = 5 ... (ii)$
$(AQ)^2= (CQ)^2$
$\Rightarrow (0 - 1)^2 + (y + 1)^2+ (z - 0)^2= (0 - 3)^2 + (y - 2)^2 (z +^1)^2$
$\Rightarrow 1 + y^2 + 2y + 1 + z^2 = 9 + y^2 - 4y + 4 + z^2+ 1 +2z$
$\Rightarrow 6y - 2z = 12 ... (iii)$
Solving equation (i) and (ii) we get
$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$
Put the value of y and z in equation (iii)
$6y - 2z = 12 = 12$
$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$
$\frac{186}{16}+\frac{6}{16}=12$
$\frac{192}{16}=12$
$12=12$
$LHS = RHS.$
so,
$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$
Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$
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Question 165 Marks
Determine the points xy-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1)$.
Answer
Let the point on xy-plane be $P(x, y, 0).$
Now P is equidistance from $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
So, $AP - BP - CP$
Now,
$(AP)^2= (x - 1)^2 + (y + 1)^2 + (0 - 0)^2$
$(BP)^2 = (x - 2)^2 + (y - 1)^2 + (0 - 2)^2$
$(CP)^2 = (x - 3)^2 + (y - 2)^2 + (0 + 1)^2$
$(AP)^2 = (BP)^2$
$\Rightarrow (x - 1)^2 + (y + 1)^2 - (x - 2)^2 + (y - 1)^2 + 4$
$\Rightarrow x^2 + 1 - 2 + y^2 + 1 + 2y + z^2 - x^2 + 4 - 4x + y^2 + 1 - 2y + 4$
$\Rightarrow 2x + 4y = 7 ... (i)$
$(BP)^2 = (CP)^2$
$\Rightarrow (x - 2)^2+ (y - 1)^2 + 4 = (x - 3)^2 + (y - 2)^2 + 1$
$\Rightarrow x^2+ 4 - 4x + y^2 + 1 - 2y + z^2 + 4 = x^2 + 9 - 6x + y^2 + 4 - 4y +1$
$\Rightarrow 2x + 2y = 5 ... (ii)$
$(AP)^2= (CP)^2$
$\Rightarrow (x - 1)^2 + (y + 1)^2+ (x - 3)^2+ (y - 2)^2+^1$
$\Rightarrow x^2+ 1 - 2x + y^2 + 1 + 2y = x^2 + 9 - 6x + y^2 + 4 - 4y + 1$
$\Rightarrow 4x + 5y = 12 ... (iii)$
Solving equation (i) and (ii) we get
y = 1, $\text{x} = \frac{3}{2}$
Put x and y in equation (iii)
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$
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Question 175 Marks
Determine the points zx-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
Answer
Let $R(x, 0, z)$ be the required point.
So
$(AR)^2 = (BR)^2 \Rightarrow (1 - x)^2 + (-1 - 0)^2 + (0 - z)^2 = (2 - x) + (1 - 0)^2 + (2 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 4 + x^2 - 4x + 1 + z^2 + 4z$
$\Rightarrow 2x + 4z = 7 ... (i)$
$(BR)^2 = (CR)^2 \Rightarrow (z - z)^2+ (1 - 0)^2 + (2 - z)^2 = (3 - x)^2 + (2 - 0)^2 (-1 - z)^2$
$\Rightarrow 4 + x^2- 4x + 4 + z^2- 4z = 9 + x^2- 6x + 4 + 1^+ z^2 + 2z$
$\Rightarrow 2x - 6z = 5 ... (ii)$
$(AR)^2= (CR)^2 \Rightarrow (1 - x)^2 + (1 - 0)^2+ (0 - z)^2= (3 - x)^2 + (2 - 0)^2 + (-1 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 9 + 6x + 4 + 1 + z^2$
$\Rightarrow 4x - 2z = 12 ... (iii)$
Solving equation (i) and (ii) we get
$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$
Put the value of x and z in equation (iii)
4x - 2z = 12
$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$
$\frac{124}{10}+\frac{2}{10}=12$
$\frac{120}{10}=12$
$12=12$
LHS = RHS.
so,
$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$
Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$
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Question 185 Marks
Using distance formula prove that the following points are collinear:
A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Answer
A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

$\text{AB}=\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}$

$=\sqrt{(4)^2+(-5)^2+(-7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

$\text{BC}=\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}$

$=\sqrt{(-8)^2+(10)^2+(14)^2}$

$=\sqrt{64+100+196}$

$=\sqrt{360}$

$=6\sqrt{10}\text{ units}$

$\text{CA}=\sqrt{(3-7)^2+(-5+10)^2+(1+6)^2}$

$=\sqrt{(-4)^2+(5)^2+(7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

Since, AB + AC = BC

so, A, B and C are collinear.
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