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

3 June, 2026 at 8:15 pm #1060

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

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