If the area of parallelogram, whose diagonals are $\hat{i}-\hat{j}+2 \hat{k}$ and $2 \hat{i}+3 \hat{j}+\alpha \hat{k}$ is $\frac{\sqrt{93}}{2}$ sq. unit, then $\alpha=$

Given the diagonals of the parallelogram $\mathrm{d_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\mathrm{d_2} = 2\hat{i} + 3\hat{j} + \alpha\hat{k}$, the area is determined by the formula $\text{Area} = \frac{1}{2} |\mathrm{d_1} \times \mathrm{d_2}|$.

The cross product evaluates to $\mathrm{d_1} \times \mathrm{d_2} = (-\alpha - 6)\hat{i} - (\alpha - 4)\hat{j} + 5\hat{k}$.

Its magnitude is $|\mathrm{d_1} \times \mathrm{d_2}| = \sqrt{(\alpha + 6)^2 + (\alpha - 4)^2 + 25} = \sqrt{2\alpha^2 + 4\alpha + 77}$.

Given the area $\frac{\sqrt{93}}{2}$, we have $\frac{1}{2} \sqrt{2\alpha^2 + 4\alpha + 77} = \frac{\sqrt{93}}{2}$, which simplifies to $\sqrt{2\alpha^2 + 4\alpha + 77} = \sqrt{93}$.

Squaring both sides yields $2\alpha^2 + 4\alpha + 77 = 93$, leading to $2\alpha^2 + 4\alpha - 16 = 0$.

Dividing by 2 gives $\alpha^2 + 2\alpha - 8 = 0$, which factors as $(\alpha + 4)(\alpha - 2) = 0$.

The solutions are $\alpha = -4$ and $\alpha = 2$, corresponding to option A.

The final answer is $\boxed{\text{A}}$.