(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

Simplify to get \(3n^3 - 6n^2 + n\).

(b) Partial fraction decomposition: \(\frac{1}{(2-3r)(5-3r)} = \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Sum: \(\sum_{r=1}^{n} \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Telescoping series: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum is \(\frac{1}{6}\).