Using cofactors of elements of third column, evaluate $\Delta =\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$

Given: $\Delta =\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$
Minor of $a_{13}=M_{13}=\left|\begin{array}{cc}1& y\\ 1& z\end{array}\right|$
$\Rightarrow M_{13}=1\times z-1\times y=z-y$
Minor of $a_{23}=M_{23}=\left|\begin{array}{cc}1& x\\ 1& z\end{array}\right|$
$\Rightarrow M_{23}=1\times z-1\times x=z-x$
Minor of $a_{33}=M_{33}=\left|\begin{array}{cc}1& x\\ 1& y\end{array}\right|$
$\Rightarrow M_{33}=1\times y-1\times x=y-x$
Cofactor of $a_{13}=A_{13}=(-1{)}^{1+3}{M}_{13}=(-1{)}^4.(z-y)=z-y$
Cofactor of $a_{23}=A_{23}=(-1{)}^{2+3}.M_{23}=(-1{)}^5.(z-x)=x-z$
Cofactor of $a_{33}=A_{33}=(-1{)}^{3+3}.M_{33}=(-1{)}^6.(y-x)=y-x$
Now,
$\Delta =a_{13}{A}_{13}+a_{23}{A}_{23}+a_{33}{A}_{33}$
$\Rightarrow \Delta =yz(z-y)+zx(x-z)+xy(y-x)$
$\Rightarrow \Delta =y{z}^2-y^{2}z+z{x}^2-z^{2}x+x{y}^2-x^{2}y$
$\Rightarrow \Delta =(y{z}^2-y^{2}z)+(x{y}^2-z^{2}x)+(z{x}^2-x^{2}y)$
$\Rightarrow \Delta =yz(z-y)+x(y^2-z^2)+x^2(z-y)$
$\Rightarrow \Delta =-yz(y-z)+x(y^2-z^2)-x^2(y-z)$
$\Rightarrow \Delta =-yz(y-z)+x(y+z)(y-z)-x^2(y-z)$
$\Rightarrow \Delta =(y-z)(-yz+x(y+z)-x^2)$
$\Rightarrow \Delta =(y-z)\left(z\right(x-y)-x(x-y\left)\right)$
$\Rightarrow \Delta =(x-y)(y-z)(z-x)$
Therefore, the value of determinant is $(x-y)(y-z)(z-x)$