(1,2,0)
(1,0,5)
(0,3,9)
(1,2,5)
Let the equation of the plane be ax+by+cz+d=0, where a,b,c are direction ratios of normal to the plane. The plane contains the line x=y/2=z/3. Hence, it passes through the point (1,2,3) and is parallel to the line. ⇒a(x-1)+b(y-2)+c(z-3)=0 and a+2b+3c=0 The given plane is parallel to x=y=z/4 ⇒a+b+4c=0 Eliminating a,b,c from the above equations gives | x y z | | 1 2 3 | = 0 ⇒5x-y-z=0 | 1 1 4 | Clearly, (1,0,5) lies on the plane 5x-y-z=0 Hence, option B is correct.