JEE Main 2026 — 8 April, Evening Shift — Question 32

$$\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$$

$\mathrm{Q}(1, \mu, 2)$ satisfies the line L $\therefore \frac{1-4}{1}=\frac{\mu-9}{2}=\frac{2-5}{1}$ $\Rightarrow-3=\frac{\mu-9}{2}=-3 \Rightarrow \mu=3$ D.R. of PQ : $\lambda-1,2-\mu, 1$

Since PQ $\perp \mathrm{L} \Rightarrow(\lambda-1) .1+2(2-\mu)+1.1=0$ $\Rightarrow \lambda-2 \mu+4=0$ $\therefore \lambda=2$

Now lines : $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ & $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z+5}{6}$ i.e. $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+t_{1}(2 \hat{i}+3 \hat{j}+6 \hat{k})$ $\& \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})+\mathrm{t}_{2}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$ Distance $=\frac{|((\hat{i}+2 \hat{j}-4 \hat{k})-(2 \hat{i}+3 \hat{j}-5 \hat{k})) \times(2 \hat{i}+3 \hat{j}+6 \hat{k})|}{|2 \hat{i}+3 \hat{j}+6 \hat{k}|}$ $=\frac{|-9 \hat{i}+8 \hat{j}-\hat{k}|}{7}=\frac{\sqrt{146}}{7}$