A) 1
B) 2
C) 4
D) none of these
Correct Answer: A
Solution :
[a] : The equation of the line passing through \[P\left( 1,-2,3 \right)\]and parallel to the given line is \[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}\] Suppose it meets the plane \[x-y+z=5\]at the point Q given by\[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda \] i.e.,\[(2\lambda +1,3\lambda -2,-6\lambda +3)\] This lies on x - y + z = 5. Therefore, \[2\lambda +1-3\lambda +2-6\lambda +3=5\Rightarrow -7\lambda =-1\Rightarrow \lambda =\frac{1}{7}\]So, the co-ordinates of Q are (9/7, -11/7, 15/7). Hence, required distance \[=PQ=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1\].
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