The distance from the point (3,4,5) to the point where the line x-3 1 = y-4 2 = z-5 2 meets the plane x+y+z=17 is

Step-by-step solution

Anypoint on the line is (r+3,2 r+4,2 r+5) . It lies on the plane x+y+z=17 (r+3)+(2 r+4)+(2 r+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