Let a line pass through two distinct points P(-2, -1, 3) and Q , and be parallel to the vector 3i+2j+2k .

Question

Let a line pass through two distinct points \(\mathrm{P}(-2,-1,3)\) and Q , and be parallel to the vector \(3 \hat{i}+2 \hat{j}+2 k.\) If the distance of the point Q from the point R(1, 3, 3) is 5 , then the square of the area of \(\triangle \mathrm{PQR}\) is equal to:

(1) 136

(2) 140

(3) 144

(4) 148

Answer 1

Correct option is (1) 136  

\(\overrightarrow{\mathrm{PQ}}\) parallel to \(3 \hat{i}+2 \hat{j}+2 \hat{k}, R(1,3,3)\)

\(\Rightarrow \mathrm{Q}(3 \lambda-2,2 \lambda-1,2 \lambda+3), \lambda \in \mathrm{R}-\{0\}\)

\(|\overrightarrow{\mathrm{QR}}|=5=\sqrt{(3 \lambda-3)^{2}+(2 \lambda-4)^{2}+(2 \lambda)^{2}}\)

\(\therefore 17 \lambda^{2}-34 \lambda+25=25 \Rightarrow \lambda=2(\because \lambda \neq 0)\)

\(\therefore \mathrm{Q}(4,3,7), \mathrm{P}(-2,-1,3), \mathrm{R}(1,3,3)\)

Area of \(\triangle \mathrm{PQR}=[\mathrm{PQR}]=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}|\)

\([\mathrm{PQR}]=\frac{1}{2}\left\|\begin{array}{lll}\hat{i} & \hat{\mathrm{j}} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0\end{array}\right\|=\left\|\begin{array}{lll}\hat{i} & \hat{\mathrm{j}} & \hat{k} \\ 3 & 2 & 2 \\ 3 & 4 & 0\end{array}\right\|\)

\([\mathrm{PQR}]=|-8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}|=\sqrt{136}\)

\(\therefore[\mathrm{PQR}]^{2}=136\)