BITSAT 2020 Paper

Anypoint on the line is (r+3,2r+4,2r+5).
It lies on the plane x+y+z=17
∴(r+3)+(2r+4)+(2r+5)=17 i.e r=1
Thus the point of intersection of the plane and the line is (4,6,7)
Required distance = distance between (3,4,5) and (4,6,7) =√{(4−3)2+(6−4)2+(7−5)2}=3