$$
\begin{array}{l}
\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, b=\mathbf{i}+3 \mathbf{j}+5 \mathbf{k} \\
\text { and } \mathbf{c}=7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k} \\
\text { Let } \mathbf{A}=\mathbf{a}+\mathbf{b} \\
=(\mathbf{i}+\mathbf{j}+\mathbf{k})+(\mathbf{i}+3 \mathbf{j}+5 \mathbf{k}) \\
=2 \mathbf{i}+4 \mathbf{j}+6 \mathbf{k} \\
\text { and } \mathbf{B}=\mathbf{b}+\mathbf{c} \\
=(\mathbf{i}+3 \mathbf{j}+5 \mathbf{k})+(7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k}) \\
=8 \mathbf{i}+12 \mathbf{j}+16 \mathbf{k}
\end{array}
$$
$\therefore$ Area of parallelogram
$$
\begin{array}{l}
=\frac{1}{2}|\mathbf{A} \times \mathbf{B}| \\
(\because \text { A and B are diagonals }) \\
=\frac{1}{2}\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 4 & 6 \\
8 & 12 & 16
\end{array}\right| \\
=\frac{1}{2}|\mathbf{i}(64-72)-\mathbf{j}(32-48)+\mathbf{k}(24-32)| \\
=\frac{1}{2} \vdash 8 \mathbf{i}+16 \mathbf{j}-8 \mathbf{k} \mid \\
=\sqrt{(-4)^{2}+(8)^{2}+(-4)^{2}} \\
=\sqrt{96}=4 \sqrt{6} \mathrm{sq} \text { units }
\end{array}
$$