Answer
$$9 x+10 y+5 z =33$$
Work Step by Step
Given $$x z+2 x^{2} y+y^{2} z^{3}=11, \quad P=(2,1,1)$$
Consider $f(x,y,z)=x z+2 x^{2} y+y^{2} z^{3}-11 $, since
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle z+4 x y, 2 x^{2}+2 y z^{3}, x+3 y^{2} z^{2}\right\rangle\\
\nabla f_{P}&=\langle 9,10,5\rangle
\end{align*}
Then the equation of the tangent plane at $P$ is given by
\begin{align*}
\nabla f_{P} \cdot\langle x-x_1, y-y_1, z-z_1\rangle&= 0\\
\langle 9,10,5\rangle \cdot\langle x-2, y-1, z-1\rangle&=0\\
9(x-2)+10(y-1)+5(z-1)&=0\\
9 x+10 y+5 z&=33
\end{align*}
