Equation of plane passing through (1,−3,−2) and perpendicular to planes x+2y+2z=5 and 3x+3y+2z=8 is
A
2x+4y+3z+8=0
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B
2x−4y+3z−8=0
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C
2x−4y+3z+8=0
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D
2x−4y−3z−8=0
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Solution
The correct option is B2x−4y+3z−8=0 Given planes: P1:x+2y+2z=5 and P2:3x+3y+2z=8
normal to P1:→n1=^i+2^j+2^k
normal to P2:→n2=3^i+3^j+2^k Normal vector of required plane =→n1×→n2
=∣∣
∣
∣∣^i^j^k122332∣∣
∣
∣∣=−2^i+4^j−3^k
So, Equation of plane −2x+4y−3z=d passing through (1,−3,−2)d=−8−2x+4y−3z=−82x−4y+3z−8=0