JEE Main 2025 — 29 January, Morning Shift — Question 49

$\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}$\ $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$ ,, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0$

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

$\Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$

$\overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$|\overrightarrow{\mathrm{c}}|^{2}=\lambda^{2}(25+36+16)$

$|\overrightarrow{\mathrm{c}}|^{2}=77 \lambda^{2}$

$(\vec{a}+\vec{c}) \cdot(\vec{b}+\vec{c})=168$

$\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{c} \cdot \vec{b}+|\vec{c}|^{2}=168$

$14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^{2}=168$

using equation (1) $\begin{aligned} & \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^{2}+77 \lambda^{2}=154 \\ & 77 \lambda+77 \lambda^{2}-154=0 \\ & \lambda^{2}+\lambda-2=0 \\ & \lambda=-2,1 \end{aligned}$

$\therefore$ Maximum value of $|\overrightarrow{\mathrm{c}}|^{2}$

occurs when $\lambda=-2$

$|\overrightarrow{\mathrm{c}}|^{2}=77 \lambda^{2}$

$=77 \times 4$

$=308$