Question

Let the equation of the plane, that passes through the point \((1, 4, -3)\) and contains the line of intersection of the planes \(3x - 2y + 4z - 7 = 0\) and \(x + 5y - 2z + 9 = 0\), be \(\alpha x + \beta y + \gamma z + 3 = 0\), then \(\alpha + \beta + \gamma\) is equal to :

• A. \(23\)
• B. \(15\)
• C. \(-15\)
• D. \(-23\)

Hint:

Pay attention to the constant term in the required plane equation. In this case, the standard result ended in \(-3\), so a sign change was needed to match the given form \(\dots + 3 = 0\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A plane passing through the line of intersection of two planes \(P_1 = 0\) and \(P_2 = 0\) is given by the family of planes \(P_1 + \lambda P_2 = 0\).
Step 2: Detailed Explanation:
The family of planes is:
\[ (3x - 2y + 4z - 7) + \lambda (x + 5y - 2z + 9) = 0 \]
Since the plane passes through the point \((1, 4, -3)\):
\[ [3(1) - 2(4) + 4(-3) - 7] + \lambda [1 + 5(4) - 2(-3) + 9] = 0 \]
\[ [3 - 8 - 12 - 7] + \lambda [1 + 20 + 6 + 9] = 0 \]
\[ -24 + \lambda(36) = 0 \implies \lambda = \frac{24}{36} = \frac{2}{3} \]
Substitute \(\lambda = 2/3\) back into the family equation:
\[ 3(3x - 2y + 4z - 7) + 2(x + 5y - 2z + 9) = 0 \]
\[ 9x - 6y + 12z - 21 + 2x + 10y - 4z + 18 = 0 \]
\[ 11x + 4y + 8z - 3 = 0 \]
To make the constant term \(+3\), multiply the entire equation by \(-1\):
\[ -11x - 4y - 8z + 3 = 0 \]
Comparing this with \(\alpha x + \beta y + \gamma z + 3 = 0\):
\(\alpha = -11\), \(\beta = -4\), \(\gamma = -8\).
Calculate \(\alpha + \beta + \gamma\):
\[ -11 - 4 - 8 = -23 \]
Step 3: Final Answer:
The sum \(\alpha + \beta + \gamma\) is \(-23\), which is option (D).