Answer: \(k=\frac23\).

Use gradients for straight-line work: parallel lines have equal gradients and perpendicular lines have gradients whose product is \(-1\).

The midpoint of \((-3,\frac23)\) and \((6,-\frac73)\) is

\(\left(\frac{-3+6}{2},\frac{\frac23-\frac73}{2}\right)=\left(\frac32,-\frac56\right).\)

The gradient of the line joining the two points is

\(\displaystyle \frac{-\frac73-\frac23}{6-(-3)}=\frac{-3}{9}=-\frac13.\)

Therefore the perpendicular bisector has gradient \(3\).

Using the midpoint, its equation is

\(y+\frac56=3\left(x-\frac32\right).\)

Since \((2,k)\) lies on this line,

\(k+\frac56=3\left(2-\frac32\right)=\frac32.\)

Thus

\(k=\frac32-\frac56=\frac23.\)