The equation of the plane through the point (1, 2, 3) and parallel to the plane x + 2y + 5z = 0 is

$(x-1) +2(y-2) + 5(z-3) = 0 $ $x + 2y + 5z  14$ $x + 2y + 5z = 6 $ none of these

$(x-1) +2(y-2) + 5(z-3) = 0 $

Required plane is parallel to the plane x + 2y + 5z = 0. Therefore, direction ratios of a vector normal to the plane are proportional to 1, 2, 5. Hence, the equation of the required plane is $(x-1) +2(y-2) + 5(z-3) = 0 $