JEE Mathematics 2025 Question with Solution

Given: $Q(10,-3,-1)$, $R(3,-2,1)$, and the line

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

Find: The area of right-angled triangle PQR, where P is the foot of the perpendicular from $Q$ to the given line.

A point on the line is $A(3,2,-1)$ and its direction vector is

$$\vec{a} = \langle 7,-1,-2 \rangle$$

So any point on the line is

$$P = (3+7\lambda,\, 2-\lambda,\, -1-2\lambda)$$

Since $P$ is the foot of the perpendicular from $Q$ to the line, vector $\overrightarrow{PQ}$ is perpendicular to the direction vector $\vec{a}$.

For point $P(3+7\lambda, 2-\lambda, -1-2\lambda)$,

$$\overrightarrow{PQ} = \langle 3+7\lambda-10,\, 2-\lambda+3,\, -1-2\lambda+1 \rangle$$

that is,

$$\overrightarrow{PQ} = \langle 7\lambda-7,\, -\lambda+5,\, -2\lambda \rangle$$

Apply the perpendicularity condition:

$$\langle 7\lambda-7,\, -\lambda+5,\, -2\lambda \rangle \cdot \langle 7,-1,-2 \rangle = 0$$

So,

$$49\lambda - 49 + \lambda - 5 + 4\lambda = 0$$ $$54\lambda - 54 = 0$$

Hence,

$$\lambda = 1$$

Therefore,

$$P = (3+7,\, 2-1,\, -1-2) = (10,1,-3)$$

Now,

$$\overrightarrow{PQ} = Q - P = \langle 10-10,\, -3-1,\, -1-(-3) \rangle = \langle 0,-4,2 \rangle$$

Using the working shown in the solution for the area calculation,

$$\overrightarrow{PR} = R - P = \langle 3-10,\, -2-1,\, 1-(-3) \rangle = \langle -7,-3,4 \rangle$$

The area of triangle $PQR$ is

$$\text{Area} = \frac{1}{2}\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right|$$

Compute the cross product as shown:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \langle -10,-14,28 \rangle$$

Its magnitude is

$$\sqrt{(-10)^2+(-14)^2+28^2} = \sqrt{100+196+784} = \sqrt{1080}$$

Therefore,

$$\text{Area} = \frac{\sqrt{1080}}{2} = 3\sqrt{30}$$

So, the area of the triangle is $3\sqrt{30}$, and the correct option is D.