$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3a}&{ - a + b}&{ - a + c}\\{ - b + a}&{3b}&{ - b + c}\\{ - c + a}&{ - c + b}&{3c}\end{array}} \right| = 3(a + b + c)(ab + bc + ca)$

Let $\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3a}&{ - a + b}&{ - a + c}\\{ - b + a}&{3b}&{ - b + c}\\{ - c + a}&{ - c + b}&{3c}\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$ ,

we get
$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a + b + c}&{ - a + b}&{ - a + c}\\{a + b + c}&{3b}&{ - b + c}\\{a + b + c}&{ - c + b}&{3c}\end{array}} \right|$

Taking ($a + b + c$) common from ${C_1}$,

we get
$\Delta = (a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - a + b}&{ - a + c}\\1&{3b}&{ - b + c}\\1&{ - c + b}&{3c}\end{array}} \right|$

Applying ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1},$

we get
$\Delta = (a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - a + b}&{ - a + c}\\0&{2b + a}&{a - b}\\0&{a - c}&{2c + a}\end{array}} \right|$

Expanding along ${C_1},$

we get
$\Delta = (a + b + c)[(2b + a(2c + a) - (a - c)(a - b)]$

$= (a + b + c)(4bc + 2ab + 2ac + {a^2} - {a^2} + ab + ac - bc)$

$= (a + b + c)(3ab + 3bc + 3ca) = 3(a + b + c)(ab + bc + ca)$

Hence, proved.