JEE Main 2025 — Mathematics Vectors & 3D Geometry

$\begin{aligned}
& \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \
& \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \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 \
& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{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}}) \ldots . .(1) \
& |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \
& |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \
& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \
& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168
\end{aligned}$$14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^2=168$using equation (1)$usingequation(1)\begin{aligned}
& \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^2+77 \lambda^2=154 \
& 77 \lambda+77 \lambda^2-154=0 \
& \lambda^2+\lambda-2=0 \
& \lambda=-2,1
\end{aligned}$$\therefore$Maximum value of$Maximumvalueof|\overrightarrow{\mathrm{c}}|^2$occurs when$occurswhen\lambda=-2$$\begin{aligned}
& |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \
& =77 \times 4 \
& =308
\end{aligned}$