$$
\text {Let } \mathrm{ABCD} \text { be a parallelogram such that }
$$


$$
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{q}} \Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{q}}
$$
then $\overline{\mathrm{AC}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{a}}$
Similarly, $\overline{\mathrm{BD}}=-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overline{\mathrm{b}}$
$\vec{a}+\vec{b}=2 \vec{q} \Rightarrow \vec{q}=\frac{1}{2}(\vec{a}+\vec{b})$
and $\vec{a}-\vec{b}=2 \vec{p} \Rightarrow \vec{p}=\frac{1}{2}(\vec{a}-\vec{b})$
Now, $\vec{p} \times \vec{q}=\frac{1}{4}(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})$
$$
\begin{aligned}
&=\frac{1}{4}(\vec{a} \times \vec{a}+\vec{a} \times b-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}) \\
&=\frac{1}{4}[\vec{a} \times \vec{b}+\vec{a} \times \vec{b}]=\frac{1}{2}(\vec{a} \times \vec{b})
\end{aligned}
$$
Hence, area of a parallelogram
$$
\mathrm{ABCD}=|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|=\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|
$$
Now, area of a parallelogram, whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}$.
$$
\begin{aligned}
&=\frac{1}{2}|(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})| \\
&=\frac{1}{2} \mid \begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & -1 & 1 \\
1 & 3 & -1
\end{array} \|
\end{aligned}
$$
$=\frac{1}{2}|[\hat{\mathrm{i}}(1-3)-\hat{\mathrm{j}}(-2-1)+\hat{\mathrm{k}}(6+1)]|$
$=\frac{1}{2}|-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{4+9+49}$
$=\frac{1}{2} \sqrt{62}$ sq units