JEE Main 2025 — 29 January, Morning Shift — Question 58

$\quad L_{1}: \vec{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})$

$\Rightarrow \overrightarrow{\mathrm{r}}=(\lambda-1) \hat{\mathrm{i}}+2(\lambda+1) \hat{\mathrm{j}}+(\lambda+1) \hat{\mathrm{k}}$

$L_{2}: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$\Rightarrow \overrightarrow{\mathrm{r}}=2 \mu \hat{\mathrm{i}}+(1+7 \mu) \hat{\mathrm{j}}+(1+3 \mu) \hat{\mathrm{k}}$

For point of intersection equating respective components

$\Rightarrow \lambda-1=2 \mu$

$2(\lambda+1)=1+7 \mu$

$\lambda+1=1+3 \mu$

We get $\Rightarrow \lambda=3$ and $\mu=1$

$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

$L_{3}: \vec{r}=2 \hat{i}+8 \hat{j}+4 \hat{k}+\alpha(3 \hat{i}+9 \hat{j}+4 \hat{k})$

For $\alpha=2, \overrightarrow{\mathrm{r}}=8 \hat{\mathbf{i}}+26 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}$