JEE Main 2024 — 6 April, Shift 2 — Question 14

\begin{array}{*{35}{r}}{} & \text{RQ}=\sqrt{1+64+4}=\sqrt{69} \\{} & \overrightarrow{\text{RQ}}=\overset{\text{}}{\mathop{\text{i} }}\,-8\overset{\text{}}{\mathop{\text{j}}}\,+2\overset{\text{}}{\mathop{\text{k}}}\, \\{} & \overrightarrow{\text{RS}}=\overset{\text{}}{\mathop{\text{i}}}\,+\overset{\text{}}{\mathop{\text{j}}}\,-\overset{\text{}}{\mathop{\text{k}}}\, \\{} & \text{cos}\theta =\frac{\overrightarrow{\text{RQ}}\cdot \overrightarrow{\text{RS}}}{\left| \overrightarrow{\text{RQ}} \right|\left| \overrightarrow{\text{RS}} \right|}=\left| \frac{1-8-2}{\sqrt{69}\sqrt{3}} \right|=\frac{9}{3\sqrt{23}}\\{}&\text{cos}\theta=\frac{3}{\sqrt{23}}=\frac{\text{RS}}{\text{RQ}}=\frac{\text{RS}}{\sqrt{69}} \\{} & \text{RS}=3\sqrt{3} \\{} & \text{sin}\theta =\frac{\sqrt{14}}{\sqrt{23}}=\frac{\text{QS}}{\sqrt{69}} \\{} & \text{QS}=\sqrt{42} \\{} & \text{area}=\frac{1}{2}\cdot 2\text{QS}\cdot \text{RS}=\sqrt{42}\cdot 3\sqrt{3} \\{} & \lambda =9\sqrt{14} \\{} & {{\lambda }^{2}}=81.14=14\text{k} \\{} & \boxed{{k}=81} \\\end{array}