JEE Main 2026 — 21 January, Morning Shift — Question 14

$\bar{r}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$

$\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{-1}=\lambda$

Any general point P on the line is $(2 \lambda-1,3 \lambda+3,-\lambda+1)$

Let the given point is $\mathrm{A}(5,4,2)$ $\overline{\operatorname{AP}}(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}+(-\lambda-1) \hat{\mathrm{k}}$

$\because \overline{\mathrm{AP}} \perp^{\mathrm{r}}$

Line $(\mathrm{L})$ $\therefore \overline{\mathrm{AP}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0$

$2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0$

$\Rightarrow \lambda=1$

$\therefore \alpha=1$ $\beta=6$

$\gamma=0$

Let the vector $\overline{\mathrm{u}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$

$\overline{\mathrm{u}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+O \hat{\mathrm{k}}$ & $\overline{\mathrm{w}}=6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

So projection $=\frac{|\overline{\mathrm{u}} \cdot \overline{\mathrm{w}}|}{|\overline{\mathrm{w}}|}=\frac{18}{7}$