(c) We have, ${x_1} + 2{x_2} + 3{x_3} = c$
$2a{x_1} + 3{x_2} + {x_3} = c$
$3b{x_1} + {x_2} + 2{x_3} = c$
Let $a = b = c = 1$.
Then $D = \left| {\,\begin{array}{*{20}{c}}1&2&3\\2&3&1\\3&1&2\end{array}\,} \right|$ = $1\,(5) - 2\,(1) + 3\,( - 7) = - 18 \ne 0$
${D_x} = \left| {\,\begin{array}{*{20}{c}}1&2&3\\1&3&1\\1&1&2\end{array}\,} \right| = - 3$
Similarly ${D_y} = {D_z} = - 3$. Now, $x = \frac{{{D_z}}}{D}$ = $\frac{1}{6}$
Hence $D \ne 0$, $x = y = z$, i.e., unique solution.