A) \[7x-8y+3z-25=0\]
B) \[7x-8y+3z+25=0\]
C) \[-7x+8y-3z+5=0\]
D) \[7x-8y-3z+5=0\]
Correct Answer: B
Solution :
Given, equaiton of plane is passing through the point (?1, 3, 2) \[\therefore \] \[A(x+1)+B(y-3)+C\,(z-2)=0\] .....(i) Since plane (i) is perpendicular to each of the planes \[x+2y+3z=5\] and \[3x+3y+z=0\] So, \[A+2B+3C=0\] and \[3A+3B+C=0\] \[\therefore \] \[\frac{A}{2-9}=\frac{B}{9-1}=\frac{C}{3-6}=K\] Þ \[A=-7K,\,B=8K,\,C=-3K\] Put the values of A, B and C in (i) we get, \[7x-8y+3z+25=0\], which is the required equation of the plane.You need to login to perform this action.
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