We have given two circles are
$2 x^{2}+2 y^{2}-3 x+6 y +k =0$
$\Rightarrow x^{2}+y^{2}-\frac{3}{2} x+3 y+\frac{k}{2} =0$
and $x^{2}+y^{2}-4 x+10 y+16=0$
Since, general equation of circle is
$x^{2}+y^{2}+2 g x+2 f y +c=0$
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 $g_{2}=-2, f_{2}=5, c_{2}=16$
Since, both the circles cut orthogonally.
$\therefore 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2}$
$\Rightarrow 2\left(\frac{3}{2}+\frac{15}{2}\right)=\frac{k}{2}+16$
$\Rightarrow 18=\frac{k}{2}+16$
$\Rightarrow \frac{k}{2}=2$
$\Rightarrow k=4$