Question

Let P be the point of intersection of the line \(\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}\) and the plane x + y + z = 2. If the distance of the point P from the plane 3x – 4y + 12z = 32 is q, then q and 2q are the roots of the equation 

• A. \(x^2+18x+72=0\)
• B. \(x^2-18x-72=0\)
• C. \(x^2+18x-72=0\)
• D. \(x^2-18x+72=0\)

Answer 1

The Correct Option is D

Given: \( \frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = \lambda \).

• Step 1: Parametric Equations of the Line:

\( x = 3\lambda - 3, \quad y = \lambda - 2, \quad z = 1 - 2\lambda \).

• Point \( P(3\lambda - 3, \lambda - 2, 1 - 2\lambda) \) lies on the plane \( x + y + z = 2 \). Substitute into the plane equation:

\( 3\lambda - 3 + \lambda - 2 + 1 - 2\lambda = 2 \).

• Simplify:

\( 2\lambda - 4 = 2 \implies \lambda = 3 \).

• Thus:

\( P(6, 1, -5) \).

• Step 2: Perpendicular Distance of \( P \) from the Plane:

Plane equation: \( 3x - 4y + 12z - 32 = 0 \).

• Substitute \( P(6, 1, -5) \) into the plane equation:

\( q = \frac{|3(6) - 4(1) + 12(-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}} \).

• Simplify:

\( q = \frac{|18 - 4 - 60 - 32|}{\sqrt{9 + 16 + 144}} = \frac{|-78|}{13} = 6 \).

• Step 3: Roots of the Quadratic Equation:
  • Given roots: \( q = 6 \) and \( 2q = 12 \).
  • Sum of roots: \( 6 + 12 = 18 \).
  • Product of roots: \( 6 \cdot 12 = 72 \).

Final Answer: The quadratic equation is:

\( x^2 - 18x + 72 = 0 \).