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Question 16 — JEE Main 21 Jan 2026 (S I)

Question

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\vec{r}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$. Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma k$ on the vector $6 \hat{i}+2 \hat{j}+3 k$ is
[JEE MAIN 21 Jan 2026 S I]
(1) $\frac{15}{7}$
(2) 4
(3) $\frac{18}{7}$
(4) 3

Solution

let point B on given line
$\mathrm{B}(2 \lambda-1,3 \lambda+3,1-\lambda)$
$\mathrm{AB} \xrightarrow{\mathrm{drs}} 2 \lambda-1-5,3 \lambda+3-4,1-\lambda-2$
or $2 \lambda-6,3 \lambda-1,-\lambda-1$

line $\xrightarrow{\text { drs }} 2,3,-1$
$\mathrm{AB} \perp$ line
$2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0$
$\Rightarrow 4 \lambda-12+9 \lambda-3+\lambda+1=0 \Rightarrow 14 \lambda=14 \Rightarrow \lambda=1$
$\mathrm{B}(1,6,0) \equiv(\alpha, \beta, \gamma)$
$\alpha=1, \beta=6, \gamma=0$
required length of Projection

\[ =\frac{|(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}) \cdot(6 \hat{i}+2 \hat{j}+3 \hat{k})|}{\sqrt{36+4+9}}=\left|\frac{6 \alpha+2 \beta+3 \gamma}{7}\right|=\frac{6+12+0}{7}=\frac{18}{7} \text { Answer }(3) \]