The given two circles are
$$
\begin{array}{l}
2 x^{2}+2 y^{2}-3 x+6 y+k=0...(i) \\
\Rightarrow x^{2}+y^{2}-\frac{3}{2} x+3 y+\frac{k}{2}=0...(ii)
\end{array}
$$
Since, general equation of circle is
$$
x^{2}+y^{2}+2 g x+2 f y+c=0...(iii)
$$
Therefore, comparing eqs. (i) and (ii) with eq. (iii), we get
$$
g_{1}=-\frac{3}{4}, f_{1}=\frac{3}{2}, c_{1}=\frac{k}{2}
$$
and $g_{2}=-2, f_{2}=5, c_{2}=16$
Both the circles cut orthogonally,
$$
\begin{array}{l}
\therefore 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2} \\
\Rightarrow 2\left(\frac{3}{2}+\frac{15}{2}\right)=\frac{k}{2}+16 \\
\Rightarrow 18=\frac{k}{2}+16 \Rightarrow \frac{k}{2}=2 \Rightarrow k=4
\end{array}
$$