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Question

Let L1:x13=y21=z33 be a line and P:4x+3y+5z50=0 be a plane. L2 is the line in the plane P and parallel to L1. If equation of the plane containing both the lines L1 and L2 and perpendicular to plane P is ax+by+5z+d=0, then the value of a+b+d is

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Solution

Line L1 is parallel to the plane P.
L2 is parallel to L1 and lying in the plane P.
Now, normal of the plane containing L1 and L2 both is to the plane P and the lines.

Direction ratios of normal
=∣ ∣ ∣^i^j^k313435∣ ∣ ∣

=14^i27^j+5^k
Hence, equation of the required plane is
14(x1)27(y2)+5(z3)=0
or, 14x27y+5z+25=0
a+b+d=1427+25=12

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