Given two vectors are $\hat{\mathit{i}}-2\hat{\mathit{j}}+\hat{\mathit{k}}$ and $3\hat{\mathit{i}}+2\hat{\mathit{j}}+\hat{\mathit{k}}$
⇒ sin θ = $\frac{\left|\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}\right|}{\left|\overrightarrow{\mathit{a}}\right||\overrightarrow{\mathit{b}}|}$ ... (i)
To find $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$
$\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$ = $\left|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 1& -2& 3\\ 3& -2& 1\end{array}\right|$
⇒ $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$ = [- 2 - (- 6)] $\hat{\mathit{i}}$ - (1 - 9) $\hat{\mathit{j}}$ + [- 2 - (- 6)] $\hat{\mathit{k}}$
⇒ $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$ = $4\hat{\mathit{i}}+8\hat{\mathit{j}}+4\hat{\mathit{k}}$
⇒ $\left|\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}\right|$ = $\sqrt{4^2+8^2+4^2}$ = $\sqrt{96}$ = 4 $\sqrt{6}$ ... (i)
$\left|\overrightarrow{\mathit{a}}\right|$ = $\sqrt{1^2+{\left(-2\right)}^2+3^2}$ = $\sqrt{14}$ ... (ii) and $\left|\overrightarrow{\mathit{b}}\right|$ = $\sqrt{3^2+{\left(-2\right)}^2+3^2}$ = $\sqrt{14}$ ... (iii)
Since sin θ = $\frac{\left|\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}\right|}{\left|\overrightarrow{\mathit{a}}\right||\overrightarrow{\mathit{b}}|}$
⇒ sin θ = $\frac{4\sqrt{6}}{\sqrt{14}\sqrt{14}}$ ... From (i), (ii) and (iii)
⇒ sin θ = $\frac{2\sqrt{6}}{7}$