Equation of plane passing through $A(3,-2,1)$ and perpendicular to $4 \hat{i}+7 \hat{j}-4 \hat{k}$ is
$$
\begin{aligned}
& 4(x-3)+7(y+2)-4(z-1)=0 \\
& \Rightarrow 4 x+7 y-4 z+6=0
\end{aligned}
$$
Now, $\frac{\alpha-1}{4}=\frac{\beta-2}{7}=\frac{\gamma+1}{-4}=\frac{-1(4+14+4+6)}{4^2+7^2+4^2}$
$$
\Rightarrow \frac{\alpha-1}{4}=\frac{\beta-2}{7}=\frac{\gamma+1}{-4}=\frac{-28}{81}
$$
$\begin{aligned} & P M=\sqrt{(\alpha-1)^2+(\beta-2)^2+(\gamma+1)^2} \\ & =\sqrt{4^2\left(\frac{-28}{81}\right)^2+7^2\left(\frac{-28}{81}\right)^2+(-4)^2\left(\frac{-28}{81}\right)^2} \\ & =\frac{28}{81} \sqrt{4^2+7^2+4^2}=\frac{28}{81} \times 9=\frac{28}{9} .\end{aligned}$