JEE Main 2025 — 29 January, Morning Shift — Question 61

DR's of $L_{3}=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ -1 & 2 & 1\end{array}\right|$

$=-5 \hat{i}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

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

$$\mathrm{A}(\alpha-5 \lambda, \beta-3 \lambda, \gamma+\lambda)$$

$L_{1}: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}=k$

$${\mathrm{B}(\mathrm{k}+1,-\mathrm{k}+2,2 \mathrm{k}+1)}$$

Now $\alpha-5 \lambda=k+1 \Rightarrow \alpha=5 \lambda+\mathrm{k}+1$

$\beta-3 \lambda=-k+2 \Rightarrow \beta=3 \lambda-k+2$

$\gamma+\lambda=2 \mathrm{k}-1 \Rightarrow \gamma=-\lambda+2 \mathrm{k}+1$

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

$=25$