### Three Dimensional Geometry Question 3

#### Question 3 - 2024 (01 Feb Shift 2)

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $R(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is:

(1) 26

(2) 36

(3) 18

(4) 24

## Show Answer

#### Answer (3)

#### Solution

$P(8 \lambda-3,2 \lambda+4,2 \lambda-1)$

$PR=6$

$(8 \lambda-4)^{2}+(2 \lambda+2)^{2}+(2 \lambda-4)^{2}=36$

$\lambda=0,1$

Hence $P(-3,4,-1) \& Q(5,6,1)$

Centroid of $\triangle PQR=(1,4,1) \equiv(\alpha, \beta, \gamma)$

$\alpha^{2}+\beta^{2}+\gamma^{2}=18$