The point of intersection of the lines whose parametric equations are x = 3r + 3, y = − 4r + 2, z = r − 1 and x = − 6t + 1, y = 3t − 2 and z = 1 is (A) (9, 6, −1) (B) (9, −6, 1) (C) (−9, 6, −1) (D) (9, 6, 1)

The point of intersection of the lines whose parametric equations are

1.	The point of intersection of the lines whose parametric equations are x = 3r + 3, y = − 4r + 2, z = r − 1 and x = − 6t + 1, y = 3t − 2 and z = 1 is (A) (9, 6, −1) (B) (9, −6, 1) (C) (−9, 6, −1) (D) (9, 6, 1)
Answer» Correct option is (B) (9, − 6, 1) there is one shortcut too...substitute r = 2 ...then first equation will become x = 9 , y = -6 , z = 1 respectively....and thus the answer

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The point of intersection of the lines whose parametric equations are x = 3r + 3, y = − 4r + 2, z = r − 1 and x = − 6t + 1, y = 3t − 2 and z = 1 is (A) (9, 6, −1) (B) (9, −6, 1) (C) (−9, 6, −1) (D) (9, 6, 1)

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