Irrotational means the field can be written as the gradient of a scalar potential, and equivalently its curl is zero. We compute $\nabla \times \vec{F}$ from the determinant form and force each row to vanish.
Writing the curl determinant with rows $(\hat{i},\hat{j},\hat{k})$, $(\partial_x,\partial_y,\partial_z)$, and $(F_x,F_y,F_z)$:
The $\hat{i}$ term needs $\partial_y F_z = \partial_z F_y$. Since $F_z = x - by + 7z$ gives $\partial_y F_z = -b$, and $F_y = cx - 5y - 2z$ gives $\partial_z F_y = -2$, we match $-b = -2$, so $b = 2$.
The $\hat{j}$ term needs $\partial_z F_x = \partial_x F_z$. Here $\partial_z F_x = a$ from $F_x = 3x - 4y + az$, and $\partial_x F_z = 1$. Matching gives $a = 1$.
The $\hat{k}$ term needs $\partial_x F_y = \partial_y F_x$. Here $\partial_x F_y = c$ and $\partial_y F_x = -4$. Matching gives $c = -4$.
So the ordered triple is $a = 1$, $b = 2$, $c = -4$.
\[\boxed{(a,b,c) = (1,\ 2,\ -4)}\]