$\left| {\begin{array}{cccccccccccccccccccc}{3x}&{ - x + y}&{ - x + z}\\{x - y}&{3y}&{z - y}\\{x - z}&{y - z}&{3z}\end{array}} \right|$

We have, $\left| {\begin{array}{cccccccccccccccccccc}{3x}&{ - x + y}&{ - x + z}\\{x - y}&{3y}&{z - y}\\{x - z}&{y - z}&{3z}\end{array}} \right|$

Applying, ${C_1} \to {C_1} + {C_2} + {C_3}$,
$= \left| {\begin{array}{cccccccccccccccccccc}{x + y + z}&{ - x + y}&{ - x + z}\\{x + y + z}&{3y}&{z - y}\\{x + y + z}&{y - z}&{3z}\end{array}} \right|$

$= (x + y + z)\left| {\begin{array}{cccccccccccccccccccc}1&{ - x + y}&{ - x + z}\\1&{3y}&{z - y}\\1&{y - z}&{3z}\end{array}} \right|$

[taking $(x + y + z)$ common from column ${C_1}$ ]
$= (x + y + z)\left| {\begin{array}{cccccccccccccccccccc}1&{ - x + y}&{ - x + z}\\0&{2y + x}&{x - y}\\0&{x - z}&{2z + x}\end{array}} \right|$

and $\left. {{R_3} \to {R_3} - {R_1}} \right]$
Now, expanding along first column,

we get
$(x + y + z) \cdot 1\,[(2y + x)(2z + x) - (x - y)(x - z)]$

$= (x + y + z)\left( {4yz + 2yx + 2xz + {x^2} - {x^2} + xz + yx - yz} \right)$

$= (x + y + z)(3yz + 3yx + 3xz)$

$= 3(x + y + z)(yz + yx + xz)$