For the linear transformation S compute the preimages.
\[S: C^3 \rightarrow C^3, \; \; \; \; S\left(\Biggl[\begin{array}{cc}
a \\
b \\
c
\end{array}\Biggr]\right) = \left[\begin{array}{cc}
a-2b-c \\
3a-b+2c \\
a+b+2c
\end{array}\right]\]
\[S^{-1}\left(\Biggl[\begin{array}{cc} -2 \\ 5 \\ 3 \end{array}\Biggr]\right)\; \; \; \; S^{-1}\left(\Biggl[\begin{array}{cc} -5 \\ 5 \\ 7 \end{array}\Biggr]\right)\]
\(S^{-1}\left(\Biggl[\begin{array}{cc} -2 \\ 5 \\ 3 \end{array}\Biggr]\right) = \left[\begin{array}{cc} 1 & -2 & -1 & -2\\ 3 & -1 & 2 & 5 \\ 1 & 1 & 2 & 3 \end{array}\right]\)
\[R_3 = R_3 - R_1: \left[\begin{array}{cc} 1 & -2 & -1 & -2\\ 3 & -1 & 2 & 5 \\ 0 & 3 & 3 & 5 \end{array}\right]\]
\[R_2 = R_2 - 3R_1: \left[\begin{array}{cc} 1 & -2 & -1 & -2\\ 0 & 5 & 5 & 11 \\ 0 & 3 & 3 & 5 \end{array}\right]\]
\[R_2 = \frac{1}{5}R_2: \left[\begin{array}{cc} 1 & -2 & -1 & -2\\ 0 & 1 & 1 & \frac{11}{5} \\ 0 & 3 & 3 & 5 \end{array}\right]\]
\[R_3 = R_3 - 3R_2: \left[\begin{array}{cc} 1 & -2 & -1 & -2\\ 0 & 1 & 1 & \frac{11}{5} \\ 0 & 0 & 0 & -\frac{8}{5} \end{array}\right]\]
\[R_1 = R_1 + 2R_2: \left[\begin{array}{cc} 1 & 0 & 1 & \frac{12}{5}\\ 0 & 1 & 1 & \frac{11}{5} \\ 0 & 0 & 0 & -\frac{8}{5} \end{array}\right]\]
\[T^{-1}(v) = \{u \: \epsilon \: C^3 \: | \: T(u) = v\}\]
\[ = \left\{\Biggl[\begin{array}{cc} u_1 \\ u_2 \\ u_3 \end{array}\Biggr] \: \Biggl | \: u_1 = \frac{12}{5}-u_3, \: u_2 = \frac{11}{5} - u_3 \right\} \]
\[ = \left\{\Biggl[\begin{array}{cc} \frac{12}{5}-u_3 \\ \frac{11}{5}-u_3 \\ u_3 \end{array}\Biggr] \: \Biggl | \: u_3 \: \epsilon \: C \right\} \]
\[ = \left\{\Biggl[\begin{array}{cc} \frac{12}{5} \\ \frac{11}{5} \\ 0 \end{array}\Biggr] + u_3 \Biggl[\begin{array}{cc} -1 \\ -1 \\ 1 \end{array}\Biggr]\: , \: u_3 \: \epsilon \: C \right\} \]
\[ = \Biggl[\begin{array}{cc} \frac{12}{5} \\ \frac{11}{5} \\ 0 \end{array}\Biggr] + \Biggl< \Biggl[\begin{array}{cc} -1 \\ -1 \\ 1 \end{array}\Biggr] \Biggr> \]
\(S^{-1}\left(\Biggl[\begin{array}{cc} -5 \\ 5 \\ 7 \end{array}\Biggr]\right) = \left[\begin{array}{cc} 1 & -2 & -1 & -5\\ 3 & -1 & 2 & 5 \\ 1 & 1 & 2 & 7 \end{array}\right]\)
\[R_3 = R_3 - R_1: \left[\begin{array}{cc} 1 & -2 & -1 & -5\\ 3 & -1 & 2 & 5 \\ 0 & 3 & 3 & 12 \end{array}\right]\]
\[R_2 = R_2 - 3R_1: \left[\begin{array}{cc} 1 & -2 & -1 & -5\\ 0 & 5 & 5 & 20 \\ 0 & 3 & 3 & 12 \end{array}\right]\]
\[R_3 = \frac{1}{3}R_3: \left[\begin{array}{cc} 1 & -2 & -1 & -5\\ 0 & 5 & 5 & 20 \\ 0 & 1 & 1 & 4 \end{array}\right]\]
\[R_2 = R_2 - 5R_3: \left[\begin{array}{cc} 1 & -2 & -1 & -5\\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 4 \end{array}\right]\]
\[R_1 = R_1 + 2R_3: \left[\begin{array}{cc} 1 & 0 & 1 & 3\\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 4 \end{array}\right]\]
\[T^{-1}(v) = \{u \: \epsilon \: C^3 \: | \: T(u) = v\}\]
\[ = \left\{\Biggl[\begin{array}{cc} u_1 \\ u_2 \\ u_3 \end{array}\Biggr] \: \Biggl | \: u_1 = 3-u_3, \: u_2 = 4 - u_3 \right\} \]
\[ = \left\{\Biggl[\begin{array}{cc} 3-u_3 \\ 4-u_3 \\ u_3 \end{array}\Biggr] \: \Biggl | \: u_3 \: \epsilon \: C \right\} \]
\[ = \left\{\Biggl[\begin{array}{cc} 3 \\ 4 \\ 0 \end{array}\Biggr] + u_3 \Biggl[\begin{array}{cc} -1 \\ -1 \\ 1 \end{array}\Biggr]\: , \: u_3 \: \epsilon \: C \right\} \]
\[ = \Biggl[\begin{array}{cc} 3 \\ 4 \\ 0 \end{array}\Biggr] + \Biggl< \Biggl[\begin{array}{cc} -1 \\ -1 \\ 1 \end{array}\Biggr] \Biggr> \]