d
(d) By ${C_1} \to {C_1} + {C_2} + {C_3}$,
we have $(9 + x)$ $\left| {\,\begin{array}{*{20}{c}}1&3&5\\1&{x + 2}&5\\1&3&{x + 4}\end{array}\,} \right|\, = 0$
$ \Rightarrow $ $(x + 9)$ $\left| {\,\begin{array}{*{20}{c}}0&{1 - x}&0\\0&{ - (1 - x)}&{1 - x}\\1&3&{x + 4}\end{array}\,} \right| = 0$
$ \Rightarrow $ $(x + 9)$ ${(1 - x)^2}\left| {\,\begin{array}{*{20}{c}}0&1&0\\0&{ - 1}&1\\1&3&{x + 4}\end{array}\,} \right| = 0$
$ \Rightarrow $ $x = 1,\,1,\, - 9$,