Consider the lines $\mathrm{L}_1: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and $\mathrm{L}_2:…

Consider the lines $\mathrm{L}_1: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and $\mathrm{L}_2: \mathrm{x}-2=\mathrm{y}=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $\mathrm{P}(5,1,-3)$ on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ be $Q$ and $R$ respectively. If the area of the triangle PQR is A , then $4 \mathrm{~A}^2$ is equal to :
  1. 139
  2. 147
  3. 151
  4. 143

Solution

$\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \\ & \text { Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \\ & \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{m}}=0\end{aligned}$

$\begin{aligned} \Rightarrow & \lambda-4+\lambda+1, \lambda+3=0 \\ \Rightarrow & 3 \lambda=0 \\ & \lambda=0\end{aligned}$
$\begin{aligned} & L_2: \frac{x-2}{1}=\frac{y-0}{1}=\frac{z-1}{2} \\ & \text { Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4) \\ & \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0 \\ & \mu-3+\mu-1+\mu+4=0 \\ & \neq \mu=0\end{aligned}$
$\begin{aligned} & \text { Area of } \triangle \mathrm{PQR}(\mathrm{A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \\ & \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \\ & \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \\ & \left|\begin{array}{crr}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -4 & 1 & 3 \\ -3 & -1 & 4\end{array}\right| \\ & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & 4 \mathrm{~A}^2=49 \times 3=147\end{aligned}$ .

Asked in: JEE Main 2025 (07 Apr Shift 2)

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