The line joining the points (1,1,2) and (3,−2,1) meets the plane 3x+2y+z=6 at the point.
A
(1,1,2)
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B
(3,−2,1)
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C
(2,−3,1)
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D
(3,2,1)
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Solution
The correct option is C(3,−2,1) The straight line joining the points (1,1,2) and (3,−2,1) is x−12=y−1−3=z−2−1=r(say) Therefore, point is (2r+1,1−3r,2−r) which lies on 3x+2y+z=6. Thus 3(2r+1)+2(1−3r)+2−r=6 ⇒r=1 So, required point is (3,−2,1).