Let P(3,2,6) be a point in space and Q be a point on line r→=(i^−j^+2k^)+μ(−3i^+j^+5k^). Then the value of μ for which the vector PQ→ is parallel to the plane x−4y+3z=1 is
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a
14
b
-14
c
18
d
-18
answer is A.
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Detailed Solution
Any point on the line can be taken as Q={(1−3μ),(μ−1),(5μ+2)}PQ→={−3μ−2,μ−3,5μ−4) Now, 1(−3μ−2)−4(μ−3)+3(5μ−4)=0or −3μ−2−4μ+12+15μ−12=0or 8μ=2 or μ=1/4