Решим первую систему уравнений:
\(\left\{ {\begin{array}{*{20}{c}}{{{\left( {x-6y} \right)}^{-1}} = -0,1,}\\{7x-2y = 2a\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x-6y = -10,}\\{7x-2y = 2a\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x = 6y-10,\,\,\,\;\;\;\;\;\;\;\;}\\{42y-70-2y = 2a}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x = 6y-10,}\\{y = \frac{{a + 35}}{{20}}\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x = \frac{{3a + 105}}{{10}}-10,}\\{y = \frac{{a + 35}}{{20}}\;\;\;\;\,\;\;\;\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x = \frac{{3a + 5}}{{10}},}\\{y = \frac{{a + 35}}{{20}}.}\end{array}} \right.\)
Решим вторую систему уравнений:
\(\left\{ {\begin{array}{*{20}{c}}{4x + y = 2a,\,}\\{\frac{1}{{x-4y}} = -\frac{1}{6}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{4x + y = 2a,}\\{x-4y = -6\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{y = 2a-4x,\;\;\;\,\;\;\;\;}\\{x-8a + 16x = -6}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{y = 2a-4x,}\\{x = \frac{{8a-6}}{{17}}\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{y = 2a + \frac{{24-32a}}{{17}},}\\{x = \frac{{8a-6}}{{17}}\,\,\;\;\;\;\;\;\;\;\;\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{y = \frac{{24 + 2a}}{{17}},}\\{x = \frac{{8a-6}}{{17}}.\;\;\;}\end{array}} \right.\)
Найдём, при каком значении параметра a решения систем совпадают:
\(\left\{ {\begin{array}{*{20}{c}}{\frac{{3a + 5}}{{10}} = \frac{{8a-6}}{{17}},\;\,}\\{\frac{{a + 35}}{{20}} = \frac{{24 + 2a}}{{17}}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{51a + 85 = 80a-60,\;\;\;\,}\\{17a + 595 = 480 + 40a}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{29a = 145,}\\{23a = 115\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;a = 5.\)
Ответ: \(5.\)