The field F is irrotational if curl F = 0

$\bar{V} \times F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x + 2y + az & bx - 3y - z & 4x + cy + 2z \end{vmatrix} \\ = \hat{i}\bigg[\frac{\partial}{\partial y}(4x + cy + 2z) - \frac{\partial}{\partial z}(bx - 3y - z)\bigg] + \hat{j}\bigg[\frac{\partial}{\partial x}(4x + cy + 2z) - \frac{\partial}{\partial z}(x + 2y + az)\bigg] + \hat{k}\bigg[\frac{\partial}{\partial x}(bx - 3y - z) - \frac{\partial}{\partial y}(x + 2y + az)\bigg] \\ \bar{V}\times F = \hat{i}(c + 1) + \bar{j}(4 - a) + \hat{k}(b - z) \\ \bar{V} \times F = 0 \\ \hat{i}(c + 1) + \bar{j}(4 - a) + \bar{k}(b - z) = 0 \\ \therefore c+ 1 = 0, a = 4, b = 2 \\ \therefore c = -1, a = 4, b = 2$

$\therefore \bar{F} = (x + 2y + 4z)\hat{i} + (2x - 3y - z)\hat{j} + (4x - y + 2z)\hat{k}$