The plane x+3y+13=0 passes through the line of intersection of the planes 2x−8y+4z=pand3x−5y+4z+10=0. If the plane is perpendicular to the plane3x−y−2z−4=0, then the value of p is equal to

2

5

9

3

Correct Answer: Option(d)

Solution: [d] The required plane is (2+3λ)x+(−8−5λ)y+(4+4λ)z+P−10λ=0 Compare the coefficients with the plan We get 4+4λ=0⇒λ=−1 x+3y+oz+13=0 Then we get P=3.

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