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Question

The equation of the plane through intersection of planes x+2y+3z=4 and 2x+yz=5, and perpendicular to the plane 5x+3y+6z+8=0 is

A
7x2y+3z+81=0
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B
23x+14y9z+48=0
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C
51x+15y50z+173=0
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D
5x17y+2z+23=0
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Solution

The correct option is C 51x+15y50z+173=0
Equation of plane through intersection of the two planes is,
P:(x+2y+3z4)+λ(2x+yz+5)=0 (1)
nP=(1+2λ)^i+(2+λ)^j+(3λ)^k
Direction ratios of its normal are 2λ+1,λ+2,3λ

Again, consider the plane 5x+3y+6z+8=0 (2)
Direction ratios of its normal are 5,3,6

Since plane (1) is perpendicular to plane (2),
nPn=05(1+2λ)+3(2+λ)+6(3λ)=0
7λ+29=0λ=297
Putting λ=297 in equation (1), we get
(x+2y+3z4)297(2x+yz+5)=07(x+2y+3z4)29(2x+yz+5)=07x+14y+21z2858x29y+29z145=051x+15y50z+173=0

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