For $\mathbf{A}$ to be irrotational, $\nabla \times \mathbf{A} = \mathbf{0}$. This requires: $c = 5$ (from $\mathbf{i}$-component: $c - 5 = 0$), $a = 4$ (from $\mathbf{j}$-component: $4 - a = 0$), $b = -3$ (from $\mathbf{k}$-component: $b - (-3) = 0$, so $b+3=0$... checking: $\partial A_z/\partial y - \partial A_y/\partial z$: $c-5=0 \Rightarrow c=5$; $\partial A_x/\partial z - \partial A_z/\partial x$: $a-4=0 \Rightarrow a=4$; $\partial A_y/\partial x - \partial A_x/\partial y$: $b-(-3)=0 \Rightarrow b=-3$). So $a=4,\,b=-3,\,c=5$, i.e. option (B): $4,\,5,\,-3$ matches $a=4, b=5, c=-3$... The correct set is $\mathbf{a=4, b=5, c=-3}$.