Categories > 3-d > Line and plane > The distance of the point (–1, –5, –10) from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$, is

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Question: The distance of the point (–1, –5, –10) from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$, is


1) 10
2) 11
3) 12
4) 13
Solution: Explanation: No Explanation
Line and plane

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