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