Question

The plane \( x+3y+13=0 \) passes through the line of intersection of the planes \( 2x-8y+4z=p \) and \( 3x-5y+4z+10=0 \). If the plane is perpendicular to the plane \( 3x-y-2z-4=0 \), then the value of \( p \) is equal to

• A. \(2\)
• B. \(5\)
• C. \(9\)
• D. \(3\)
• E. \(-1\)

Hint:

Always use \(P_1+\lambda P_2=0\) for line of intersection problems.

Answer 1

The Correct Option is D

Concept:
• Family of planes through intersection: \( P_1 + \lambda P_2 = 0 \)
• Perpendicular planes: dot product of normals = 0

Step 1: General plane
\[ (2x-8y+4z-p) + \lambda(3x-5y+4z+10)=0 \]

Step 2: Expand
\[ (2+3\lambda)x + (-8-5\lambda)y + (4+4\lambda)z + (-p+10\lambda)=0 \]

Step 3: Normal vector
\[ \vec{n}_1 = (2+3\lambda,\ -8-5\lambda,\ 4+4\lambda) \] Other plane normal: \[ \vec{n}_2 = (3,-1,-2) \]

Step 4: Perpendicular condition
\[ (2+3\lambda)3 + (-8-5\lambda)(-1) + (4+4\lambda)(-2)=0 \]

Step 5: Simplify
\[ 6+9\lambda +8+5\lambda -8 -8\lambda = 0 \] \[ (6+8-8) + (9\lambda+5\lambda-8\lambda)=0 \] \[ 6 + 6\lambda =0 \] \[ \lambda = -1 \]

Step 6: Substitute back
\[ (2-3)x + (-8+5)y + (4-4)z + (-p-10)=0 \] \[ (-x) + (-3y) + 0z -p -10=0 \] \[ x + 3y + p +10 =0 \]

Step 7: Compare with given plane
\[ x+3y+13=0 \] \[ p+10=13 \Rightarrow p=3 \]

Step 8: Final Answer
\[ \boxed{3} \]