JEE Main 2026 — 28 January, Morning Shift — Question 17

$\mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}-1}{1}$

$\mathrm{L}_{1}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}-\mathrm{a}}{\mathrm{b}}=\lambda$

$\mathrm{L}_{2}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}-\mathrm{a}}{\mathrm{c}}=\mu$

Let $\mathrm{A}(3 \lambda+1,4 \lambda+2, \mathrm{~b} \lambda+\mathrm{a})$

It lies on L $\therefore \frac{3 \lambda}{1}=\frac{4 \lambda+2}{2}=\frac{\mathrm{b} \lambda+\mathrm{a}-1}{1}$

$\Rightarrow \lambda=1$ and $\mathrm{a}+\mathrm{b}-1=3$

$$\begin{gathered} \Rightarrow \mathrm{A}(4,6,4), \mathrm{a}+\mathrm{b}=4 \end{gathered}$$

Let $\mathrm{B}(\mu+1,4 \mu+2, \mathrm{c} \mu+\mathrm{a})$

It also lies on L $\frac{\mu}{1}=\frac{4 \mu+2}{2}=\frac{\mathrm{c} \mu+\mathrm{a}-1}{1}$

$\Rightarrow 2 \mu=4 \mu+2$ $\Rightarrow \mu=-1$

$\mathrm{a}-\mathrm{c}-1=-1$

$\Rightarrow \mathrm{a}=\mathrm{c} \quad \ldots .\& \mathrm{~B}(0,-2,0)$

also $\mathrm{PA}=\mathrm{PB}, \mathrm{P}(1,2, \mathrm{a}), \mathrm{A}(4,6,4)$

$\Rightarrow 9+16+(\mathrm{a}-4)^{2}=1+16+\mathrm{a}^{2}$

$\Rightarrow 16+8=8 \mathrm{a}$

$\mathrm{a}=3 \quad \therefore \mathrm{c}=3, \mathrm{~b}=1$

$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=7$