If $\overline{\mathrm{a}}=\overline{\mathrm{i}}+2\overline{\mathrm{j}}+3\overline{\mathrm{k}},\overline{\mathrm{b}}=-\overline{\mathrm{i}}+2\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{c}}=\overline{\mathrm{i}}+2\overline{\mathrm{j}}-2\overline{\mathrm{k}},\overline{\mathrm{n}}$ is perpendicular to both $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, and $\mathrm{\theta }$ is the angle between $\overline{\mathrm{c}}$ and $\overline{\mathrm{n}}$ then $\mathrm{sin\theta }=$

$$\begin{gathered} \overline{\mathrm{n}}=\overline{\mathrm{a}}\times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 2& 3\\ -1& 2& 1\end{array}\right| \\ =\mathrm{i}(-4)-\mathrm{j}\left(4\right)+\mathrm{k}\left(4\right) \\ \mathrm{cos\theta }=\frac{\overline{\mathrm{c}}·\overline{\mathrm{n}}}{\left|\overline{\mathrm{c}}\right| \left|\overline{\mathrm{n}}\right|}=\left|\frac{-4-8-8}{\sqrt{16+16+16} \sqrt{1+4+4}}\right| \\ =\frac{20}{4\sqrt{3}\times 3}=\frac{5}{3\sqrt{3}} \\ \mathrm{sin\theta }=\frac{\sqrt{2}}{3\sqrt{3}} \end{gathered}$$