The given equation is equivalent to $(6a - 7)^2 + (6b - 7)^2 = 50$. Number $50$ can be represented as the sum of two squares as $25 + 25$ or $1 + 49$. Hence both $6a - 7$ and $6b - 7$ must be among the numbers $7, 5, 1, -1, -5$ and $-7$. As both $a$ and $b$ are integers, only $5, -1$ and $-7$ fit. We obtain the following cases: $6a - 7 = 5$, $6b - 7 = 5$; $6a - 7 = -1$, $6b - 7 = -7$; $6a - 7 = -7$, $6b - 7 = -1$. The corresponding solutions are $a = 2, b = 2$; $a = 1, b = 0$; and $a = 0, b = 1$.

Consider the equation as a quadratic equation w.r.t. $b$. In order to have solutions, its discriminant must be non-negative, i.e., $49 - 12(3a^2 - 7a + 4) \ge 0$. This condition is equivalent to the quadratic inequality $36a^2 - 84a - 1 \le 0$, whose solutions are $(7 - \sqrt{50})/6 \le a \le (7 + \sqrt{50})/6$. As $a$ is an integer, its only suitable values are $0, 1$, and $2$, the corresponding values of $b$ are $1, 0$, and $2$.