Let PQR be a triangle with $\mathrm{R}(-1,4,2)$. Suppose $\mathrm{M}(2,1,2)$ is the mid point of PQ. The distance of the centroid of $\triangle \mathrm{PQR}$ from the point of intersection of the line $\frac{\mathrm{x}-2}{0}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}+3}{-1}$ and $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3}{-3}=\frac{\mathrm{z}+1}{1}$ is

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Three Dimensional Geometry

Let PQR be a triangle with

R (- 1, 4, 2)

. Suppose

M (2, 1, 2)

is the mid point of PQ. The distance of the centroid of

△ PQR

from the point of intersection of the line

\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}

and

\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}

Answer: (C)

R (- 1, 4, 2)

M (2, 1, 2)

is the midpoint of PQ, so

P + Q = 2 M

, and the centroid

G = \frac{P + Q + R}{3} = \frac{2 M + R}{3} = \frac{( 4 , 2 , 4 ) + ( - 1 , 4 , 2 )}{3} = (1, 2, 2)

Line 1:

\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}

gives points

(2, 2 t, - 3 - t)

.

Line 2:

\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}

gives points

(1 + s, - 3 - 3 s, - 1 + s)

.

Setting

x

-coordinates equal:

2 = 1 + s ⟹ s = 1

, giving the point

(2, - 6, 0)

; this also satisfies line 1 (with

t = - 3

), confirming the intersection point is

(2, - 6, 0)

Distance = (2 - 1)^{2} + (- 6 - 2)^{2} + (0 - 2)^{2} = 1 + 64 + 4 = 69

Key confirmed: C.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option