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Question

A plane which is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4 passes through (1, -2, 1). 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
22
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Solution

The correct option is D 22
The plane is to the planes 2x2y+z+2=0 and xy+2z=4 normal vector to the plane is given by
∣ ∣ ∣^i^j^k221112∣ ∣ ∣=^i(4+1)^j(41)+^k(2+2)=3^i3^j
and the plane passes through (1,2,1)
Equation of plane is
3(x1)3(y+2)+0(z1)=03x+33y6=03x+3y+3=0x+y+1=0
Distance of (1,2,2) from the plane x+y+1=0 is
1+2+11+1=42=22units

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