JEE Main 2025 — 3 April, Morning Shift — Question 42

$\vec{a} \times \vec{d}-\vec{b} \times \vec{d}=0$

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

$\vec{d}=t(\vec{a}-\vec{b})$

$\overrightarrow{\mathrm{d}}=\mathrm{t}(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})$

$|\overrightarrow{\mathrm{d}}|=1$

$|\mathrm{t}|=\frac{1}{3}$

$\vec{c} \cdot \vec{a}=0$

$\lambda+\mu=0$

$\mu=-\lambda$

$\overrightarrow{\mathrm{c}}=\lambda(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \quad|\overrightarrow{\mathrm{c}}|^{2}=2 \lambda^{2}$

$\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1$

$\mathrm{t}(-2,-1,2) . \lambda(0,1,-1)=1$

$\lambda \mathrm{t}=\frac{-1}{3} \Rightarrow \lambda^{2}=1$

$|3 \lambda \hat{\mathrm{~d}}+\mu \overrightarrow{\mathrm{c}}|^{2}=9 \lambda^{2}|\hat{\mathrm{~d}}|^{2}+\mu^{2}|\overrightarrow{\mathrm{c}}|^{2}+6 \lambda \mu(\hat{\mathrm{~d}} \cdot \overrightarrow{\mathrm{c}})$

$=3 \lambda^{2}+2 \lambda^{4}$ $=5$