JEE Main 2024 — 5 April, Shift 1 — Question 28

$(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$

$(2 \vec{b}+4 \vec{a}) \times \vec{c}=0$

$\overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}})=\lambda(8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+30 \hat{\mathrm{k}})$ $\vec{a} \cdot \vec{c}=130$

$8 \lambda+42 \lambda+210 \lambda=130$

$\lambda=\frac{1}{2}$

$\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+15 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=8+7+15=30$