因为 \({S_n} = \frac{{3{n^2} + n}}{2}\left( {n \in {N^*}} \right)\)，所以 \({S_{n - 1}} = \frac{{3{{(n - 1)}^2} + n - 1}}{2} = \frac{{3{n^2} - 5n + 2}}{2}(n \geqslant 2)\)，
所以 \({a_n} = {S_n} - {S_{n - 1}} = \frac{{3{n^2} + n}}{2} - \frac{{3{n^2} - 5n + 2}}{2} = 3n - 1,(n \geqslant 2)\)，
又 \({a_1} = {S_1} = \frac{{3 \times 1 + 1}}{2} = 2\) 满足上式，
所以 \({a_n} = 3n - 1,\left( {n \in {N^*}} \right)\)，
所以 \(\frac{1}{{{a_n}{a_{n + 1}}}} = \frac{1}{{(3n - 1)(3n + 2)}} = \frac{1}{3}\left( {\frac{1}{{3n - 1}} - \frac{1}{{3n{\rm{ + }}2}}} \right)\)，
所以数列 \(\left\{ {\frac{1}{{{a_n}{a_{n + 1}}}}} \right\}\) 的前10项和为 \(\frac{1}{3}\left( {\frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{{29}} - \frac{1}{{32}}} \right) = \frac{1}{3} \times \left( {\frac{1}{2} - \frac{1}{{32}}} \right) = \frac{5}{{32}}\)，
故答案为：\(\frac{5}{{32}}\)
