JEE Main 2025 Apr 3 Shift 1, Mathematics Q23: Vector Algebra

$\vec{a}\times \hat{d}=\vec{b}\times \hat{d}$

$\Rightarrow (\vec{a}-\vec{b})\times \hat{d}=0$

$\Rightarrow \hat{d}\parallel (\vec{a}-\vec{b})=(-2,-1,2)$

Since $|\vec{a}-\vec{b}|=3$,

$\hat{d}=\pm \left(-\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)$

Also, $\vec{c}\perp \vec{a}$

$\Rightarrow \lambda +\mu =0...(1)$

Given, $\vec{c}\cdot \hat{d}=1$.

$-\frac{\lambda }{3}+\frac{2\mu }{3}=1$

Using (1), $\mu =-\lambda$.

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

$\therefore \vec{c}=(0,-1,1)$

$3\lambda \hat{d}+\mu \vec{c}=-3\hat{d}+\vec{c}=(2,1,-2)+(0,-1,1)=(2,0,-1)$

$\therefore |3\lambda \hat{d}+\mu \vec{c}{|}^2=2^2+0^2+(-1{)}^2=\boxed{5}$