Answer
$$4x+12y+8z=40$$
Work Step by Step
Given $$x^{2}+3 y^{2}+4 z^{2}=20, \quad P=(2,2,1)$$
Consider $f(x,y,z)=x^{2}+3 y^{2}+4 z^{2}-20$, 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 2x,6y,8z\right\rangle\\
\nabla f_{P}&=\langle 4,12,8\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 4,12,8\rangle \cdot\langle x-2, y-2, z-1\rangle&=0\\
4(x-2)+12(y-2)+8(z-1)&=0\\
4x+12y+8z&=40
\end{align*}
