A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
Consider a point\[(r+3,\text{ }2r+4,\text{ }2r+5)\]on the given line. This point lies on the plane\[x+y+z=17,\]then \[r+3+2r+4+2r+5=17\] \[\Rightarrow \] \[5r=17-12=5\Rightarrow r=1\] Thus, intersection point of the line and the plane is (4, 6, 7). Now, distance between the points (3, 4, 5) and (4, 6, 7) is \[\sqrt{1+4+4}=3\]You need to login to perform this action.
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