Question 29 — JEE Main 28 Jan 2026 S Ii

Solution

line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}=\lambda$
$\mathrm{M}(\lambda+1,2 \lambda, \lambda+1)$
$\mathrm{PM} \xrightarrow{\mathrm{drs}} \lambda-2,2 \lambda-2, \lambda$
line $\xrightarrow{\text { drs }} 1,2,1$
$\mathrm{PM} \perp$ line $\Rightarrow \lambda-2+4 \lambda-4+\lambda=0$
$\Rightarrow 6 \lambda=6 \Rightarrow \lambda=1$
we get $\mathrm{M}(2,2,2)$ and $\mathrm{Q}(1,2,3)$
Now distance of Q from another line

we get $\mathrm{M}(2,2,2)$ and $\mathrm{Q}(1,2,3)$
Now distance of Q from another line

$\mathrm{QR} \perp$ line
$(3 t+8) 3+2(2 t+7)-2(-2 t+2)=0$
$\Rightarrow 17 \mathrm{t}+24+14-4=0$
$\Rightarrow \mathrm{t}=-2, \mathrm{R}(3,5,9)$
$\mathrm{QR} \equiv \sqrt{4+9+36}=7$

Second Method:

$\overrightarrow{\mathrm{AQ}}=-8 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\overrightarrow{\operatorname{AR}} \|(3 \hat{i}+2 \hat{j}-2 \hat{k})$

$|\overrightarrow{\mathrm{QR}}|=|\overrightarrow{\mathrm{AQ}}| \sin \theta=\frac{|\overrightarrow{\mathrm{AQ}} \times \overrightarrow{\mathrm{AR}}|}{|\overrightarrow{\mathrm{AR}}|}=|\overrightarrow{\mathrm{AQ}} \times \mathrm{AR}|$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ \frac{3}{\sqrt{17}} & \frac{2}{\sqrt{17}} & -\frac{2}{\sqrt{17}}\end{array}\right|=\frac{1}{\sqrt{17}}|18 \hat{i}-22 \hat{j}+5 \hat{k}|$
$=\sqrt{\frac{324+484+25}{17}}=\sqrt{\frac{833}{17}}=\sqrt{49}=7$

Third Method

$\overrightarrow{\mathrm{AQ}}=-8 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} ; \mathrm{AQ}=\sqrt{64+49+4}=\sqrt{117}$
$\mathrm{AR}=$ Projection of $\overrightarrow{\mathrm{AQ}}$ on given line whose $\operatorname{drs} 3,2,-2$
$=\left|\frac{-8(3)-7(2)-2(-2)}{\sqrt{9+4+4}}\right|=\frac{|-24-14+4|}{\sqrt{17}}=2 \sqrt{17}$

$\mathrm{QR}=\sqrt{\mathrm{AQ}^{2}-\mathrm{AR}^{2}}=\sqrt{117-4 \times 17}=\sqrt{49}=7$ Answer(3)