The distance of the point (−1,2,−2) from the line of intersection of the planes 2x+3y+2z=0 and x−2y+z=0 is
A
52
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B
√342
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C
√422
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D
1√2
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Solution
The correct option is B√342
DRs of line of intersection ∣∣
∣
∣∣^i^j^k2321−21∣∣
∣
∣∣=7^i−0^j−7^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=z−1=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 ⇒2r−1=0 ⇒r=12 M(12,0,−12) PM=√94+4+94=√342