The distance between the point (3, 4, 5) and the point where the line x - $\frac{3}{1}$ = y - $\frac{4}{2}$ = z - $\frac{5}{2}$ meets the plane x + y + z = 17 is

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3 \text{ The coordinates of any point on the given line are of the form } x - $\frac{3}{1}$ = y - $\frac{4}{2}$ = z - $\frac{5}{2}$ = \lambda \Rightarrow x = \lambda + 3; y = 2\lambda + 4; z = 2\lambda + 5 \text{ So, the coordiantes of the point on the given line are } \left( \lambda + 3, 2\lambda + 4, 2\lambda + 5 \right). \text{ This point lies on the plane } , x + y + z = 17 \Rightarrow \lambda + 3 + 2\lambda + 4 + 2\lambda + 5 = 17 \Rightarrow 5\lambda = 5 \Rightarrow \lambda = 1 \text{ So, the coordinates of the point are} \left( \lambda + 3, 2\lambda + 4, 2\lambda + 5 \right) = \left( 1 + 3, 2 \left( 1 \right) + 4, 2 \left( 1 \right) + 5 \right) = \left( 4, 6, 7 \right) \text{ Now, the distance between the points} \left( 4, 6, 7 \right)\text{ and } \left( 3, 4, 5 \right)\text{ is } \sqrt{\left( 3 - 4 \right)^2 + \left( 4 - 6 \right)^2 + \left( 5 - 7 \right)^2} = $\sqrt$(1 + 4 + 4) = 3 \text{ units } So, the answer is (c).

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