Answer: \(y=-\dfrac32x+\dfrac{39}{4}\), or \(6x+4y=39\).

The gradient of \(AB\) is

\(\frac{-4-2}{7-3}=-\frac32.\)

Since \(CD\) is perpendicular to \(AB\), the gradient of \(CD\) is

\(\frac23.\)

Using \(C(2,-3)\) and \(D(k,3)\),

\(\frac{3-(-3)}{k-2}=\frac23.\)

So

\(\frac6{k-2}=\frac23,\qquad k=11.\)

Thus \(D=(11,3)\). The midpoint of \(CD\) is

\(\left(\frac{2+11}{2},\frac{-3+3}{2}\right)=\left(\frac{13}{2},0\right).\)

The perpendicular bisector of \(CD\) has gradient \(-\frac32\). Therefore its equation is

\(y-0=-\frac32\left(x-\frac{13}{2}\right).\)

Hence

\(\boxed{y=-\frac32x+\frac{39}{4}},\)

or equivalently \(6x+4y=39\).