Question

If equation of the plane that contains the point \((-2,3,5)\) and is perpendicular to each of the planes \( 2x + 4y + 5z = 8 \) and \( 3x - 2y + 3z = 5 \), is \( \alpha x + \beta y + \gamma z = 97 \), then \( \alpha + \beta + \gamma \) is:

• A. 15
• B. 18
• C. 17
• D. 16

Hint:

To solve for the unknowns when the plane is perpendicular to given planes, use the dot product condition for perpendicularity, and solve the resulting system of equations.

Answer 1

The Correct Option is A

The equation of the plane through \((-2, 3, 5)\) is: \[ a(x^2) + b(y - 3) + c(z - 5) = 0 \] This plane is perpendicular to the given planes. Thus, the direction ratios of the given planes are: \[ \text{Plane 1: } 2x + 4y + 5z = 8 \quad \text{Direction ratios: } (2, 4, 5) \] \[ \text{Plane 2: } 3x - 2y + 3z = 5 \quad \text{Direction ratios: } (3, -2, 3) \] Now, using the condition of perpendicularity, we form the system of equations: \[ 2a + 4b + 5c = 0 \] \[ 3a - 2b + 3c = 0 \] We solve this system using matrix methods. The determinant of the matrix formed by the coefficients is: \[ \begin{vmatrix} 2 & 4 & 5
3 & -2 & 3
-4 & -3 & 2 \end{vmatrix} = -16 \] Now, the equation of the plane is: \[ \text{Equation of plane: } 22x + 9y + 9z - 16z = 5 \] Simplifying this: \[ \text{Equation of plane: } 2x + y + z = 16 \] Comparing this with the given equation \( \alpha x + \beta y + \gamma z = 97 \), we get: \[ \alpha = 2, \quad \beta = 1, \quad \gamma = 6 \] Thus, the value of \( \alpha + \beta + \gamma \) is: \[ 2 + 1 + 6 = 15 \]