d
(d) We have, $\left[ {\begin{array}{*{20}{c}}1&1&1\\1&{ - 2}&{ - 2}\\1&3&1\end{array}} \right]\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$
$x + y + z = 0$ ......$(i)$
$x - 2y - 2z = 3$ ......$(ii)$
$x + 3y + z = 4$.....$(iii)$
On solving $x = 1,\,y = 2,\,z = - 3$
i.e., $\left[ \begin{array}{l}{\rm{ }}1\\{\rm{ }}2\\ - 3\end{array} \right]$.