Vector Algebra

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