(a) \(3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
\(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\)
\(\therefore 3\sqrt{75} - \sqrt{12} + \sqrt{108} \equiv 15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3}\)
= \((15 - 2 + 6)\sqrt{3}\)
= \(19\sqrt{3}\).
(b) \(124_{n} = 232_{five}\)
\(124_{n} = (1 \times n^{2}) + (2 \times n^{1}) + (4 \times n^{0}) \)
= \(n^{2} + 2n + 4\)
\(232_{five} = (2 \times 5^{2}) + (3 \times 5^{1}) + (2 \times 5^{0})\)
= \(50 + 15 + 2\)
= \(67\)
\(\implies n^{2} + 2n + 4 = 67\)
\(n^{2} + 2n + 4 - 67 = 0\)
\(n^{2} + 2n - 63 = 0\)
\(n^{2} - 7n + 9n - 63 = 0\)
\(n(n - 7) + 9(n - 7) = 0\)
\((n + 9)(n - 7) = 0\)
\(\text{n = -9 or 7}\)
Hence, \(n = 7\)
\(124_{seven} = 232_{five}\)