If \(\mathrm{A}(3,2,-1), \mathrm{B}(-2,2,-3)\) and \(\mathrm{D}(-2,5,-4)\) are the vertices of a parallelogram, then the area of the parallelogram is

We have \(\mathrm{A} \equiv(3,2,-1), \mathrm{B} \equiv(-2,2,-3)\) and \(\mathrm{D} \equiv(-2,5,-4)\)
\(
\therefore \overline{\mathrm{AB}}=-5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AD}}=-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}
\)
Area of parallelogram \(=|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}|\)
Now
\(\overline{\text{AB}} \times \overline{\text{AD}}=\left|\begin{array}{ccc}\hat{\text i } & \hat{\text j } & \hat{\text k } \\ -5 & 0 & -2 \\ -5 & 3 & -3\end{array}\right|=\hat{\text i }(6)-\hat{\text j }(5)+\hat{\text k }(-15)=\) \(6 \hat{\text i }-5 \hat{\text j }-15 \hat{\text k }\)
\( \therefore \text {Area }=\sqrt{(6)^2+(-5)^2+(-15)^2}\)