If the diagonal of a parallelogram are given by $3\mathrm{i}+\mathrm{j}-2\mathrm{k},\mathrm{i}-3\mathrm{j}+4\mathrm{k}$ then the length of its sides are

$$\begin{gathered} \overline{AC}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=3\overline{\mathrm{i}}+\overline{\mathrm{j}}-2\overline{\mathrm{k}} \\ \overrightarrow{BD}=-\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\overline{\mathrm{i}}-3\overline{\mathrm{j}}+4\overline{\mathrm{k}} \\ 2\overline{\mathrm{BC}}=4\overline{\mathrm{i}}-2\overline{\mathrm{j}}+2\overline{\mathrm{k}},\Rightarrow \overline{\mathrm{BC}}=2\overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}\therefore \left|\overline{\mathrm{BC}}\right|=\sqrt{6} \\ and 2\overline{\mathrm{AB}}=2\overline{\mathrm{i}}+4\overline{\mathrm{j}}-6\overline{\mathrm{k}}\Rightarrow \overline{\mathrm{AB}}=\overline{\mathrm{i}}+2\overline{\mathrm{j}}-3\overline{\mathrm{k}}\therefore \left|\overline{\mathrm{AB}}\right|=\sqrt{14} \end{gathered}$$