$\left| {\begin{array}{cccccccccccccccccccc}{y + z}&z&y\\z&{z + x}&x\\y&x&{x + y}\end{array}} \right| = 4xyz$

We have to prove,
$\left| {\begin{array}{cccccccccccccccccccc}{y + z}&z&y\\z&{z + x}&x\\y&x&{x + y}\end{array}} \right| = 4xyz$

$\therefore$ ${\rm{LHS}} = \left| {\begin{array}{cccccccccccccccccccc}{y + z}&z&y\\z&{z + x}&x\\y&x&{x + y}\end{array}} \right|$

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

$= 2\left| {\begin{array}{cccccccccccccccccccc}{(y + z)}&z&y\\{(z + x)}&{z + x}&x\\{(x + y)}&x&{x + y}\end{array}} \right|$

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

$= 2\left| {\begin{array}{cccccccccccccccccccc}0&{z - x}&{ - x}\\0&{z + x}&x\\y&x&{x + y}\end{array}} \right|$

$= 2\left[ {y\left( {xz - {x^2} + xz + {x^2}} \right)} \right]$

$= 4xyz = {\rm{RHS}}$