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 L₁ : x - y + 3z = 6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L₁ intersects the plane L₂ : 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 \(\sqrt6\)
• B. The coordinates of R are (1, 6, 3)
• C. The centroid of the triangle PQR is \((\frac{4}{3},\frac{14}{3},\frac{5}{3})\)
• D. The perimeter of the triangle PQR is \(\sqrt2+\sqrt6+\sqrt{11}\)

Answer 1

The Correct Option is A, C

The parametric form of the line passing through \( P \) is: \[ x = r + 1, \quad y = 2r + 3, \quad z = r + 2. \] 

Finding Intersection with \( L_1 \)

Given \( L_1 : x - y + 3z = 6 \), substituting \( x, y, z \): \[ (r + 1) - (2r + 3) + 3(r + 2) = 6. \] Solving for \( r \): \[ r = 1. \] Thus, point \( Q \) is: \[ Q = (2,5,3). \]

Equation of Line Through \( Q \) Perpendicular to \( L_1 \)

The line passing through \( Q \) and perpendicular to \( L_1 \) is: \[ \frac{x - 2}{2} = \frac{y - 5}{-1} = \frac{z - 3}{3}. \]

Finding Intersection with \( L_2 \)

Given \( L_2 : 2x - y + z = -4 \), substituting \( x, y, z \): \[ 2(2 + 2\lambda) - (5 - \lambda) + (3 + 3\lambda) = -4. \] Solving for \( \lambda \): \[ \lambda = -1. \] Thus, point \( R \) is: \[ R = (1,6,0). \]

Distance \( PQ \)

Using the distance formula: \[ PQ = \sqrt{(2 - 1)^2 + (5 - 3)^2 + (3 - 2)^2} = \sqrt{6}. \]

Centroid of \( \triangle PQR \)

The centroid is given by: \[ \left(\frac{1+2+1}{3}, \frac{3+5+6}{3}, \frac{2+3+0}{3} \right) = \left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3} \right). \]

Perimeter of \( \triangle PQR \)

The perimeter is: \[ \sqrt{6} + \sqrt{11} + \sqrt{13}. \]