Question 12 — JEE Main 24 Jan 2026 S Ii

Solution

$2 \vec{a} \times \vec{c}+3 \vec{b} \times \vec{c}=0 \Rightarrow(2 \vec{a}+3 \vec{b}) \times \vec{c}=0$
$\Rightarrow \vec{c} \| 2 \vec{a}+\overrightarrow{3 b} \Rightarrow \vec{c}=\lambda(2 \vec{a}+3 \vec{b})$
$\Rightarrow \overrightarrow{\mathrm{c}}=\lambda(2(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})+3(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}))$
$\Rightarrow \overrightarrow{\mathrm{c}}=\lambda(7 \hat{\mathrm{i}}-13 \hat{\mathrm{j}}+19 \hat{\mathrm{k}})$
$(\vec{a}-\vec{b}) \cdot \vec{c}=(\hat{i}-4 \hat{j}+2 \hat{k}) \cdot \lambda(7 \hat{i}-13 \hat{j}+19 \hat{k})=-97$
$\Rightarrow \lambda(7+52+38)=-97 \Rightarrow \lambda=-1$
$\overrightarrow{\mathrm{c}}=-7 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-19 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}} \times \hat{\mathrm{k}}=-7 \hat{\mathrm{i}} \times \hat{\mathrm{k}}+13 \hat{\mathrm{j}} \times \hat{\mathrm{k}}-19 \hat{\mathrm{k}} \times \hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{c}} \times \hat{\mathrm{k}}=7 \hat{\mathrm{j}}+13 \hat{\mathrm{i}}-0$
$|\overrightarrow{\mathrm{c}} \times \hat{\mathrm{k}}|^{2}=49+169=218$ Answer(3)