by ${R_1} \to {R_1} + {R_2} + {R_3}$

$= (x + y + z)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\x&z&x\\x&y&z\end{array}\,} \right|$ ;                          by ${C_1} \to {C_1} - {C_2}$

$= (x + y + z)\,.\,\{ ({z^2} - xy) - (xz - {x^2}) + (xy - xz)\} $

$= (x + y + z)\,{(x - z)^2}$ $ \Rightarrow $ $k = 1$.

Trick : Put $x = 1,\,y = 2$, $z = 3$, then

$\left| {\,\begin{array}{*{20}{c}}5&1&2\\4&3&1\\3&2&3\end{array}\,} \right|\, = \,5(7)\, - 1(12 - 3) + 2(8 - 9)$

= $35 - 9 - 2 = 24$ and $(x + y + z)\,{(x - z)^2} = (6)\,{( - 2)^2} = 24$

$\therefore \,\,k = \frac{{24}}{{24}} = 1$.