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