JEE Mathematics 2025 Question with Solution

Given: the solution rewrites the lines in parametric form as

$$L_1 : (x,y,z) = (1,2,1) + \lambda(1,-1,2)$$

and

$$L_2 : (x,y,z) = (-1,2,0) + s(-1,2,1)$$

So the direction vectors are

$$d_1 = (1,-1,2), \qquad d_2 = (-1,2,1)$$

Find: the value of $\left|5\alpha - 11\beta - 8\gamma\right|$ as concluded in the solution.

Since $L_3$ is perpendicular to both $L_1$ and $L_2$, its direction vector is the cross product:

$$d_3 = d_1 \times d_2$$

Using the determinant expansion shown in the solution,

$$\begin{aligned} d_3 &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} \\ &= \hat{i}((-1)(1) - 2(2)) - \hat{j}((1)(1) - 2(-1)) + \hat{k}((1)(2) - (-1)(-1)) \\ &= \hat{i}(-5) - \hat{j}(3) + \hat{k}(1) \end{aligned}$$

Hence,

$$d_3 = (-5,-3,1)$$

Therefore the line $L_3$ through $({\alpha},{\beta},{\gamma})$ is written as

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

or equivalently,

$$x = \alpha - 5t, \qquad y = \beta - 3t, \qquad z = \gamma + t$$

The solution explicitly concludes:

$$\left| 5\alpha - 11\beta - 8\gamma \right| = 25$$

A later approach also boxes the same result. Since this is a numerical value answer, the required answer is $25$.

Note: the question and the solution are inconsistent in the line equations and in the asked expression, but the solution consistently gives the final numerical result as $25$. Therefore the answer recorded from the solution is 25.