# If the lines and are perpendicular, find the value of k and hence find the equation of the plane containing these lines.

The direction ratio of the line is r1 = (– 3, – 2k, 2)

The direction ratio of the line is r2 = (k, 1, 5)

Since the line and are perpendicular so

r1.r2 = 0

(– 3, – 2k, 2). (k, 1, 5) = 0

– 3k – 2k + 10 = 0

– 5k = – 10

k = 2 the equation of the line are and The equation of the plane containing the perpendicular lines and is (– 20 – 2)x – y(– 15 – 4) + z(– 3 + 8) = d

– 22x + 19y + 5z = d

The line pass through the point (1, 2, 3) so putting x = 1, y = 2, z = 3 in the equation – 22x + 19y + 5z = d we get

– 22(1) + 19(2) + 5(3) = d

d = – 22 + 38 + 15

d = 31 The equation of the plane containing the lines is – 22x + 19y + 5z = 31

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