\begin{flalign*} a) &1 + 3 \cdot \left( \frac{7}{6} - x \right) - 2(x - \frac{2}{3}) = 3 \cdot \left( \frac{1}{2} - x \right) + \frac{4}{3} \\ &\frac{3}{2} - \frac{7}{6}x + \frac{2}{3} = \frac{1}{2} - \frac{3}{2}x + \frac{4}{3} \\ &\frac{5}{6}x + \frac{4}{3} - \frac{7}{6}x = \frac{4}{6} - \frac{3}{2}x \\ &\frac{x}{3} + \frac{4}{3} = \frac{2}{3} - \frac{x}{2} \\ &\frac{x}{6} = -1 \\ &x = \boxed{-6} \end{flalign*} \begin{flalign*} b) &\frac{1}{4} \cdot \left( \frac{2}{3}x + 2 \right) - \frac{1}{2} \cdot \left( \frac{1}{2} - \frac{5}{6}x \right) = \frac{7x + 3}{12} \\ &\frac{2x}{12} + \frac{1}{2} - \frac{1}{2} + \frac{5}{12}x = \frac{7x + 3}{12} \\ &\frac{5}{12}x + \frac{1}{2} = \frac{7x + 3}{12} \\ &\frac{x}{12} = \frac{1}{2} \\ &x = \boxed{6} \end{flalign*} \begin{flalign*} c) &\frac{2x - 5}{5} - \frac{1}{10} \cdot \left( \frac{7 - 4x}{3} - \frac{3x + 13}{2} \right) = \frac{5(x + 1) - 4}{6} \\ &\frac{4x - 20}{20} - \frac{7 - 4x}{30} + \frac{3x + 13}{60} = \frac{5x + 5 - 4}{6} \\ &\frac{2x}{15} + \frac{37}{60} = \frac{5x}{12} - \frac{5}{12} \\ &\frac{x}{30} = -\frac{2}{15} \\ &x = \boxed{-2} \end{flalign*}