d
(d) Let $A = \left| {\,\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + a}\\{x + 4}&{x + 5}&{x + b}\\{x + 6}&{x + 7}&{x + c}\end{array}\,} \right|$
Applying ${C_2} \to {C_2} - {C_1},$ we get,
$A = \,\left| {\,\begin{array}{*{20}{c}}{x + 2}&1&{x + a}\\{x + 4}&1&{x + b}\\{x + 6}&1&{x + c}\end{array}\,} \right|$
${R_Z}→ {R_2} - R_1$ and ${R_3} → {R_3} - {R_1} $
$A = \left| {\,\begin{array}{*{20}{c}}{x + 2}&1&{x + a}\\2&0&{b - a}\\4&0&{c - a}\end{array}\,} \right|\, = \, - 1\,(2c - 2a - 4b + 4a)$
= $2(2b - c - a)$
$\because$ $a, b, c $ are in $A = 0.$