JEE Main 2026 — 22 January, Morning Shift — Question 11

Point $\mathrm{M} \equiv\left(3, \frac{\mathrm{~b}}{2}+1, \frac{\mathrm{c}+\mathrm{a}}{2}\right)$

satisfies the line $\frac{3-6}{3}=\frac{\frac{\mathrm{b}}{2}+1-7}{2}=\frac{\frac{\mathrm{c}+\mathrm{a}}{2}-7}{-2}$

$$\begin{gathered} -1=\frac{\mathrm{b}-12}{4}=\frac{\mathrm{c}+\mathrm{a}-14}{-4} \end{gathered}$$

$\Rightarrow \mathrm{b}=8 \quad$ & $\mathrm{c}+\mathrm{a}=18$

Now PQ ⟂ L $\Rightarrow(4 \mathrm{i}+(\mathrm{b}-2) \mathrm{j}+(\mathrm{c}-\mathrm{a}) \mathrm{k}) \cdot(3 \mathrm{i}+2 \mathrm{j}-2 \mathrm{k})=0$

$\Rightarrow 12+2(\mathrm{~b}-2)-2(\mathrm{c}-\mathrm{a})=0$ $\Rightarrow 6+(\mathrm{b}-2)-(\mathrm{c}-\mathrm{a})=0$

$\Rightarrow \mathrm{b}-\mathrm{c}+\mathrm{a}+4=0$

$\Rightarrow 8-\mathrm{c}+\mathrm{a}+4=0$

$$\begin{gathered} \Rightarrow \mathrm{c}+\mathrm{a}=12 \end{gathered}$$

From (2) & (3)

$\mathrm{c}=15 \& \mathrm{a}=3$

So $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=9+64+225=298$