If $\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k},\mathbf{b}=\mathbf{i}+3\mathbf{j}+5\mathbf{k}$ and  $\mathbf{c}=7\mathbf{i}+9\mathbf{j}+11\mathbf{k}$, then the area of the parallelogram having diagonals $\mathbf{a}+\mathbf{b}$ and $\mathbf{b}+\mathbf{c}$ is: 

Area of the parallelogram with diagonals$\mathbf{a}+\mathbf{b}$ and $\mathbf{b}+\mathbf{c}=\frac{1}{2}\left|\right(\mathbf{a}+\mathbf{b})\times (\mathbf{b}+\mathbf{c}\left)\right|$

$\begin{array}{l}=\frac{1}{2}\left|\right\{(\mathbf{i}+\mathbf{j}+\mathbf{k})+(\mathbf{i}+3\mathbf{j}+5\mathbf{k})\}\times \{(\mathbf{i}+3\mathbf{j}+5\mathbf{k})+(7\mathbf{i}+9\mathbf{j}+11\mathbf{k})\left\}\right|\\ =\frac{1}{2}\left|\right\{(2\mathbf{i}+4\mathbf{j}+6\mathbf{k})\times (8\mathbf{i}+12\mathbf{j}+16\mathbf{k})\left\}\right|\\ =4\left|\right(\mathbf{i}+2\mathbf{j}+3\mathbf{k})\times (2\mathbf{i}+3\mathbf{j}+4\mathbf{k}\left)\right|\\ =4\left|\begin{array}{lll}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 3\\ 2& 3& 4\end{array}\right||=4|-\mathbf{i}+2\mathbf{j}-\mathbf{k}\mid =4\sqrt{6}\end{array}$