JEE Main 2026 — 2 April, Morning Shift — Question 31

Line $\left(\mathrm{L}_{1}\right)$ $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}+\mathrm{a}}{5}=\frac{\mathrm{z}+\mathrm{b}+1}{7}=\mathrm{r}_{1}$ ⇒ General point P on $\mathrm{L}_{1}$ is $\left(3 \mathrm{r}_{1}-1,5 \mathrm{r}_{1}-\mathrm{a}, 7 \mathrm{r}_{1}-\mathrm{b}-1\right)$ Line ( $\mathrm{L}_{2}$ ) $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-\mathrm{b}}{4}=\frac{\mathrm{z}-2 \mathrm{a}}{7}=\mathrm{r}_{2}$ ⇒ General point Q on $\mathrm{L}_{2}$ is $\left(\mathrm{r}_{2}+2,4 \mathrm{r}_{2}+\mathrm{b}, 7 \mathrm{r}_{2}+2 \mathrm{a}\right)$ For point of intersection

3 \mathrm{r}_{1}-1=\mathrm{r}_{2}+2 \Rightarrow \mathrm{r}_{2}=3 \mathrm{r}_{1}-3 \end{gathered}$$ $$\begin{gathered} 5 \mathrm{r}_{1}-\mathrm{a}=4 \mathrm{r}_{2}+\mathrm{b} \end{gathered}$$ $$\begin{gathered} 7 \mathrm{r}_{1}-\mathrm{b}-1=7 \mathrm{r}_{2}+2 \mathrm{a} \end{gathered}$$ Since point lies on XY plane $\Rightarrow \mathrm{z}-$ coordinate $=0$ From $\mathrm{L}_{1}: 7 \mathrm{r}_{1}-\mathrm{b}-1=0 \Rightarrow 7 \mathrm{r}_{1}=\mathrm{b}+1$ From $\mathrm{L}_{2}: 7 \mathrm{r}_{2}+2 \mathrm{a}=0 \Rightarrow 2 \mathrm{a}=-7 \mathrm{r}_{2}$ Substitute $\mathrm{r}_{2}=3 \mathrm{r}_{1}-3$ $2 \mathrm{a}=-7\left(3 \mathrm{r}_{1}-3\right)$ $\Rightarrow \mathrm{a}=\frac{-2 \mathrm{lr}_{1}+21}{2}$ Put in equation (2) $5 \mathrm{r}_{1}-\mathrm{a}=4 \mathrm{r}_{2}+\mathrm{b}$ $5 \mathrm{r}_{1}-\left(\frac{-21 \mathrm{r}_{1}+21}{2}\right)=4\left(3 \mathrm{r}_{1}-3\right)+\left(7 \mathrm{r}_{1}-1\right)$ $\Rightarrow \mathrm{r}_{1}=\frac{5}{7}$ $\Rightarrow \mathrm{b}=7 \mathrm{r}_{1}-1=4$ $\Rightarrow \mathrm{a}=3 $ $\therefore \mathrm{a}+\mathrm{b}=7$