11. Solve $\frac{3 k}{2 k-5}=\frac{7}{3}$ for $k$.

Step 1 ：Cross multiply to get rid of the fractions in the equation \(\frac{3 k}{2 k-5}=\frac{7}{3}\).

Step 2 ：Solve for k in the resulting equation.

Step 3 ：The solution to the equation is \(k = \boxed{7}\). This means that when \(k = 7\), the equation \(\frac{3 k}{2 k-5}=\frac{7}{3}\) holds true.