Mathematics - Vector Question with Solution | TestHub

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