JEE Main 2026 — 22 January, Evening Shift — Question 10

$M$ is the point of intersection of $L_{1}$ & $L_{2}$

$\Rightarrow 2 \lambda-1=2 \mu-1,3 \lambda-1=3 \mu-1,6 \lambda-3=9$

$\Rightarrow \lambda=2=\mu$

$\Rightarrow \mathrm{M}(3,5,9)$

Now let point P be $(2 \mathrm{~K}-1,3 \mathrm{~K}-1,6 \mathrm{~K}-3)$ on $\mathrm{L}_{2}$

such that $\mathrm{PM}=7$

$\Rightarrow \sqrt{(2 \mathrm{~K}-4)^{2}+(3 \mathrm{~K}-6)^{2}+(6 \mathrm{~K}-12)^{2}}=7$

$\Rightarrow 49 \mathrm{~K}^{2}+196-196 \mathrm{~K}=49$

$\Rightarrow \mathrm{K}^{2}+4-4 \mathrm{~K}=1$

$\Rightarrow \mathrm{K}^{2}-4 \mathrm{~K}+3=0$

$\Rightarrow \mathrm{K}=1,3$

So points $\mathrm{P} \& \mathrm{Q}$ are $(1,2,3) \&(5,8,15)$

So sum of all co-ordinates of $\mathrm{P} \& \mathrm{Q}=34$