Question

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 \( L_1 : x - y + 3z = 6 \) at the point \( Q \). Another straight line passing through \( Q \) and perpendicular to the plane \( L_1 \) intersects the plane \( L_2 : 2x - y + z = -4 \) at the point \( R \). Then which of the following statements is (are) TRUE?

• A. The length of the line segment \( PQ \) is \( \sqrt{6} \).
• B. The coordinates of \( R \) are \( (1, 6, 0) \).
• C. The centroid of the triangle \( PQR \) is \( \left(\frac{4}{3},
• D.

Hint:

Always parameterize the line equations and substitute them into the plane equations to find the points of intersection.

Answer 1

The Correct Option is A

Equation of the line passing through \( P \) is: \[ x = r + 1, \quad y = 2r + 3, \quad z = r + 2. \] Substituting into \( L_1: x - y + 3z = 6 \): \[ (r+1) - (2r+3) + 3(r+2) = 6 \quad \Rightarrow \quad r = 1. \] Thus, \( Q = (2, 5, 3) \). Equation of the line passing through \( Q \) and perpendicular to \( L_1 \): \[ \frac{x-2}{2} = \frac{y-5}{-1} = \frac{z-3}{3}. \] Substituting into \( L_2: 2x - y + z = -4 \): \[ 2(2 + 2\lambda) - (5 - \lambda) + (3 + 3\lambda) = -4 \quad \Rightarrow \quad \lambda = -1. \] Thus, \( R = (1, 6, 0) \). Distance \( PQ \): \[ PQ = \sqrt{(2-1)^2 + (5-3)^2 + (3-2)^2} = \sqrt{6}. \] Centroid of \( \triangle PQR \): \[ \text{Centroid} = \left(\frac{1+2+1}{3}, \frac{3+5+6}{3}, \frac{2+3+0}{3}\right) = \left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right). \] Perimeter of \( \triangle PQR \): \[ \sqrt{6} + \sqrt{11} + \sqrt{13}. \]