A) \[10\]
B) \[11\]
C) \[12\]
D) \[13\]
Correct Answer: D
Solution :
Given line is\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t\](say) Then, any point on this line is \[(3t+2,\,\,4t-1,\,\,12t+2)\]. This point lies on the plane\[x-y+z=5\]. \[\therefore \]\[3t+2-4t+1+12t+2=5\] \[\Rightarrow \] \[11t=0\Rightarrow t=0\] The point is\[(2,\,\,-1,\,\,2)\]. Its distance from\[(-1,\,\,-5,\,\,-10)\]. \[=\sqrt{{{(2+1)}^{2}}+{{(-1+5)}^{2}}+{{(2+10)}^{2}}}\] \[=\sqrt{9+16+144}=13\]You need to login to perform this action.
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