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Question

The distance of the point (1,2,2) from the line of intersection of the planes 2x+3y+2z=0 and x2y+z=0 is

A
52
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B
342
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C
422
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D
12
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Solution

The correct option is B 342


DRs of line of intersection
∣ ∣ ∣^i^j^k232121∣ ∣ ∣=7^i0^j7^k
DRs =(1,0,1)
Both the given planes pass through (0,0,0), so line of intersection also passes through (0,0,0).
Equation of line of intersection
x1=y0=z1=r
M(r,0,r)
Direction ratios of PM=(r+1,2,r+2)
PM is perpendicular to the line of intersection
1(r+1)+0(2)1(r+2)=0
2r1=0
r=12
M(12,0,12)
PM=94+4+94=342

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