A) 13
B) 12
C) 11
D) 8
E) 7
Correct Answer: A
Solution :
Given line is\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=k(say)\] Any point on the line is\[(3k+2,4k-1,12k+2)\] This point lies on the plane \[x-y+z=5\] \[\therefore \]\[3k+2-(4k-1)+12k+2=5\] \[\Rightarrow \]\[11k=0\]\[\Rightarrow \] \[k=0\] \[\therefore \]Intersection point is\[(2,-1,2)\]. \[\therefore \]Distance, between points\[(2,-1,2)\]and \[(-1,-5,-10)\] \[=\sqrt{{{(-1-2)}^{2}}+{{(-5+1)}^{2}}+{{(-10-2)}^{2}}}\] \[=\sqrt{9+16+144}=13\]You need to login to perform this action.
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