Write $1$ as $\sin^2 \alpha + \cos^2 \alpha$ etc. to get
$\left| {\,\begin{array}{*{20}{c}}{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }&{\cos \beta \cos \alpha  + \sin \beta \sin \alpha }&{\cos \gamma \cos \alpha  + \sin \gamma \sin \alpha }\\{\cos \alpha \cos \beta  + \sin \alpha \sin \beta }&{{{\cos }^2}\beta  + {{\sin }^2}\beta }&{\cos \gamma \cos \beta  + \sin \gamma \sin \beta }\\{\cos \alpha \cos \gamma  + \sin \alpha \sin \gamma }&{\cos \beta \cos \gamma  + \sin \beta \sin \gamma }&{{{\sin}^2}\gamma  + {{\cos }^2}\gamma }\end{array}\,} \right|$

can be factorized into $2$ determinant

$\left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&x\\{\cos \beta}&{\sin \beta }&x\\{\cos \gamma }&{\sin \gamma }&x\end{array}\,} \right| \left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\cos \beta }&{\cos \gamma }\\{\sin \alpha }&{\sin \beta }&{\sin \gamma }\\x&x&x \end{array}\,} \right| = 0$

Alternatively: $\alpha – \beta = x $; $\beta – \gamma = y$ ;$ \gamma – \alpha = z$ $\Rightarrow x + y + z = 0$ Now expand]