JEE Main 2024 — 1 February, Shift 1 — Question 29

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\left(\mathrm{DR}^{\prime}\right.$ s of $\left.\mathrm{L}_{1}\right)$

$\overrightarrow{\mathrm{d}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\left(\mathrm{DR}^{\prime}\right.$ s of $\left.L_{2}\right)$

$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ 1 & -1 & 1\\ 1 & 1 & -1\end{array}\right|$

$=0 \hat{i}+2 \hat{j}+2 \hat{k}$

(DR's of Line perpendicular to $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ )

DR of $A B$ line $=(0,2,2)=(3+\mu-\lambda, 3+\mu+\lambda, 3-\mu-\lambda)$

$\frac{3+\mu-\lambda}{0}=\frac{3+\mu+\lambda}{2}=\frac{3-\mu-\lambda}{2}$

Solving above equation we get $\mu=-\frac{3}{2}$ and $\lambda=\frac{3}{2}$

point $\mathrm{A}=\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)$

$\mathrm{B}=\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)$

Point of $\mathrm{AB}=\left(\frac{5}{2}, 2,6\right)=(\alpha, \beta, \gamma)$

$2(\alpha+\beta+\gamma)=5+4+12=21$