A plane passes through (1,−2,1) and is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4. The distance of the plane from the point (1,2,2) is
A
0
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B
1
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C
√2
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D
2√2
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Solution
The correct option is D2√2 The equation of a plane passing through (1,−2,1) is a(x−1)+b(y+2)+c(z−1)=0 .........(i) It is perpendicular to 2x−2y+z=0 and x−y+2z=4. Therefore, 2a−2b+c=0 and a−b+2c=0 ⇒a−4+1=−b4−1=c−2+2 ⇒a−3=b−3=c0 On substituting the values of a,b and c in equation (i), we get −3(x−1)−3(y+2)+0(z−1)=0 ⇒x+y+1=0 Distance of this plane from (1,2,2) is given by, d=1+2+1√1+1=2√2