Answer
$Q(1,-1,1)$ is the closest point to $P$.
Work Step by Step
We find:
\[
(x-1)^{2}+(y+1)^{2}+(z-1)^{2}=f(x, y, z)
\]
To find the minimum, we use Lagrange Multipliers:
\[
\begin{aligned}
\nabla f &=\lambda \nabla g \Rightarrow 2(x-1) \mathbf{i}+2(y+2) \mathbf{j}+2(z-1) \mathbf{k}=\lambda(4 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \\
& \Rightarrow \lambda=\frac{x-1}{2}=\frac{2}{3}(y+1)=2(z-1) \\
& \Rightarrow x=\frac{4}{3}(y+1)+1 \quad z=\frac{1}{3}(y+1)+1
\end{aligned}
\]
So:
\[
\begin{aligned}
4 x+3 y+z &=2 \Rightarrow 4\left(\frac{4}{3}(y+1)+1\right)+3 y+\frac{1}{3}(y+1)+1=2 \\
& \Rightarrow \frac{16}{3} y+3 y+\frac{1}{3} y=2-4-\frac{16}{3}-\frac{1}{3}-1 \\
y &=-1 \\
\text { since } y=-1, \quad & x=\frac{4}{3}(-1+1)+1=1 \\
&=\frac{1}{3}(-1+1)+1=1
\end{aligned}
\]
$Q(1,-1,1)$ is the closest point to $P$. Note that the given point is on the given plane, so the distance is $0=d$.
