Solution:

(a) The distance of the point $(0,0,0)$ from the plane $3 x-4 y+12=3 \Rightarrow$ $3 x-4 y+12 z-3=0$ is
$\begin{array}{l}
\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} \\
=\frac{|3(0)-4(0)+12(0)-3|}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}} \\
=\frac{|3|}{\sqrt{9+16+144}}=\frac{3}{\sqrt{169}}=\frac{3}{13}
\end{array}$

(b) The length of perpendicular from the point $(3,-2,1)$ on the plane $2 x-y+2 z+3=0$ is
$\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$=\frac{|2(3)-(-2)+2(1)+3|}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}$
$=\frac{|13|}{\sqrt{4+1+4}}=\frac{13}{\sqrt{9}}=\frac{13}{3}$