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JEE Advanced · Mathematics · 31. 3D Geometry

A straight line drawn from the point \(P(1,3,2)\), parallel to the line \(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}\), intersects the plane \(L_1: x-y+3 z=6\) at the point \(Q\). Another straight line which passes through \(Q\) and is perpendicular to the plane \(L_1\) intersects the plane \(L_2: 2 x-y+z=-4\) at the point \(R\). Then which of the following statements is (are) TRUE?

  1. A The length of the line segment \(P Q\) is \(\sqrt{6}\)
  2. B The coordinates of \(R\) are \((1,6,3)\)
  3. C The centroid of the triangle \(P Q R\) is \(\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)\)
  4. D The perimeter of the triangle \(P Q R\) is \(\sqrt{2}+\sqrt{6}+\sqrt{11}\)
Verified Solution

Answer & Solution

Correct Answer

(C) The centroid of the triangle \(P Q R\) is \(\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { line }: \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} \\ & (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2)\end{aligned}\)
Put in \(L_1: x-y+3 z=6\)
\(\begin{aligned} & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3)\end{aligned}\)
line : \(\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}\)
\((x, y, z)=(t+2,5-t, 3 t+3)\)
Put in \(L_2: 2 x-y+z=-4\)
\(\begin{aligned} & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 \Rightarrow t=-1 \\ & R=(1,6,0)\end{aligned}\)

\(\begin{aligned} & \text { Perimeter }=\sqrt{6}+\sqrt{13}+\sqrt{11} \\ & \text { Centroid }=\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)\end{aligned}\)
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