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The distance between the point - 1 , - 5 , - 10 and the point of intersection of the line x - 2 3 = y + 1 4 = z - 2 12 with the plane x - y + z = 5 is 13 t , then t equals to

Correct answer

1

Step-by-step solution

Any point on the line is 3 r + 2 ,   4 r   – 1 ,   12 r + 2 If it lies on the given plane, then 3 r + 2 - 4 r - 1 + 12 r + 2 = 5 ⇒   11 r = 0   ⇒   r = 0 ∴   Co-ordinates of the point of intersection of the given line and given plane are 2 ,   - 1 ,   2 and thus the required distance is 2 + 1 2 + - 1 + 5 2 + 2 + 10 2 = 13 ⇒ 13 = 13 t   ⇒   t = 1

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