JEE Main 2025 — 23 January, Morning Shift — Question 9

Given

$$Q(10,-3,-1), \quad R(3,-2,1)$$

and the line

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

Direction vector of the line:

$$\vec d=\langle 7,-1,-2\rangle$$

Since $P$ is the foot of the perpendicular from $Q$ to the line,

$$PQ \perp QR$$

(because $QR$ is parallel to the given line).

From perpendicularity, the foot is

$$P=(10,1,-3)$$

Length $PQ$:

$$PQ=\sqrt{(10-10)^2+(1+3)^2+(-3+1)^2} =\sqrt{20}$$

Length $QR$:

$$QR=\sqrt{(3-10)^2+(-2+3)^2+(1+1)^2} =\sqrt{54}=3\sqrt6$$

Area of right angled triangle $PQR$:

$$\text{Area}=\frac12 \times PQ \times QR =\frac12 \times \sqrt{20} \times 3\sqrt6 =3\sqrt{30}$$ $$\boxed{3\sqrt{30}}$$