Another tangent plane to a level surface

Find an equation for the plane tangent to the surface

at the point (1,1,1).

The gradient vector \nabla f is perpendicular to the level surface at a given point, meaning it can be the equivalent of the normal vector in the formula for the tangent plane. In this problem I found \nabla f = \langle3yx^2+z^3,x^3+3zy^2,y^3+xz^2\rangle. Plugging in the point (1,1,1) got \nabla f(1,1,1)=\langle4,4,4\rangle. Therefore the tangent plane would be 4(x-1)+4(y-1)+4(z-1)=0. Is this correct?