JEE Mathematics 2025 Question with Solution

Given: The lines are

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

and

$$\frac{x - 3}{5} = \frac{y}{-1} = \frac{z - 1}{1}$$

Find: $A^2$, where $A$ is the area of the triangle formed by the lines.

From the solution working, write the lines in vector form as

$$(x, y, z) = (-2, 3, 2) + t(-3, 3, 1)$$

and

$$(x, y, z) = (3, 0, 1) + s(5, -1, 1)$$

So the direction vectors are

$$\vec{v_1} = (-3, 3, 1), \qquad \vec{v_2} = (5, -1, 1)$$

The area of the triangle formed by these two vectors is

$$A = \frac{1}{2}\left|\vec{v_1} \times \vec{v_2}\right|$$

Now,

$$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\-3 & 3 & 1 \\5 & -1 & 1 \end{vmatrix}$$

Expanding,

$$\vec{v_1} \times \vec{v_2} = 4\hat{i} + 8\hat{j} - 12\hat{k}$$

Therefore,

$$\left|\vec{v_1} \times \vec{v_2}\right| = \sqrt{4^2 + 8^2 + (-12)^2} = \sqrt{224} = 2\sqrt{56}$$

Hence,

$$A = \frac{1}{2} \times 2\sqrt{56} = \sqrt{56}$$

So,

$$A^2 = 56$$

Therefore, the required value is $56$.

Given: The solution working identifies three intersection points:

• $P = (-2, 1, 0)$
• $Q = (0, 3, 2)$
• $R = (3, 0, 1)$

Find: $A^2$ for triangle $PQR$.

Form the side vectors:

$$\overrightarrow{PQ} = (0+2, 3-1, 2-0) = (2, 2, 2)$$ $$\overrightarrow{PR} = (3+2, 0-1, 1-0) = (5, -1, 1)$$

Area of triangle $PQR$ is

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

Now compute the cross product:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\2 & 2 & 2 \\5 & -1 & 1 \end{vmatrix}$$

This gives a vector whose magnitude leads to

$$A = \sqrt{56}$$

Hence,

$$A^2 = 56$$

Therefore, the required answer is $56$.

Note: The first extracted approach contains inconsistent line labels, but both solution approaches conclude the same final value.