(i)$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right| = {(a + b + c)^3}$
(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + y + 2z}&x&y\\z&{y + z + 2x}&y\\z&x&{z + x + 2y}\end{array}} \right| = 2{(x + y + z)^3}$

(i) L.H.S. $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|$

Applying ${R_1} \to {R_1} + {R_2} + {R_3},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a + b + c}&{a + b + c}&{a + b + c}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}} \right|$

Taking $(a + b + c)$ common from ${R_1},$

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

Applying ${C_1} \to {C_1} - {C_2}$

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

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

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

Applying ${C_2} \to {C_2} - {C_3},$

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

Taking $(a + b + c)$ common from ${C_2}$,

we get
${(a + b + c)^3}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\1&{ - 1}&{2b}\\0&1&{c - a - b}\end{array}} \right|$

Expanding along ${R_1},$

we get
${(a + b + c)^3}$ $=$ R.H.S.
Hence, proved.

(ii)

L.H.S.$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + y + 2z}&x&y\\z&{y + z + 2x}&y\\z&x&{z + x + 2y}\end{array}} \right|$

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

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{(x + y + z)}&{ - (x + y + z)}&0\\0&{x + y + z}&{ - (x + y + z)}\\z&x&{z + x + 2y}\end{array}} \right|$

Taking $(x + y + z)$ common from ${R_1}$ and ${R_2},$

we get
${(x + y + z)^2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\0&1&{ - 1}\\z&x&{z + x + 2y}\end{array}} \right|$

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

we get
${(x + y + z)^2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&{ - 1}\\z&{x + z}&{z + x + 2y}\end{array}} \right|$

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

we get
${(x + y + z)^2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\z&{x + z}&{2(x + y + z)}\end{array}} \right|$

$= 2{(x + y + z)^2} \cdot (x + y + z)$

[Determinant of triangular matrix is product of its diagonal elements]

$=$ $2{(x + y + z)^3} =$ R.H.S.
Hence, proved.