A) (2, 1, 0)
B) (7, ?1, ?7)
C) (1, 2, ?6)
D) (5, ?1, 1)
Correct Answer: D
Solution :
Point on the line, \[\frac{x-6}{-1}=\frac{y+1}{0}=\frac{z+3}{4}=r\] are \[(-\,r+6,\,\,-1,\,\,4r-3)\] This will be satisfy plane \[x+y-z=3\] \[\therefore \,\,-r+6-1-4r+3=3\,\,\Rightarrow \,\,-5r+5=0\]\[\Rightarrow \,\,r=1\] Required co-ordinates of point \[\equiv \,(5,\,\,-1,\,\,1)\].
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