= Solution
Adding the three equations for $s,t$ gives the necessary <solvability condition> $a+b+c=3$. Conversely, subtraction of the first two and use of the third force
$$
s=\frac{a-b}{2},\qquad t=\frac{1-c}{2}.
$$
If $a+b+c=3$, these values also satisfy the first two equations: for example,
$$
1+s+t=\frac{3+a-b-c}{2}=a.
$$
The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:
$$
\boxed{a+b+c=3,\qquad s=\frac{a-b}{2},\quad t=\frac{1-c}{2}.}
$$
For the three-variable system the <matrix> and data vector are
$$
A=\begin{pmatrix}5&2&-1\\2&5&-1\\-1&-1&8\end{pmatrix},
\qquad
\mathbf x=\begin{pmatrix}x\\y\\z\end{pmatrix},
\qquad
\mathbf b=\begin{pmatrix}1+s+t\\1-s+t\\1-2t\end{pmatrix},
\qquad A\mathbf x=\mathbf b.
$$
Perform <Gaussian elimination> on the augmented matrix. The operations $R_2\leftarrow5R_2-2R_1$ and $R_3\leftarrow5R_3+R_1$ give
$$
\left[
\begin{array}{ccc|c}
5&2&-1&1+s+t\\
0&21&-3&3-7s+3t\\
0&-3&39&6+s-9t
\end{array}
\right].
$$
Each operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow7R_3+R_2$ gives
$$
\left[
\begin{array}{ccc|c}
5&2&-1&1+s+t\\
0&21&-3&3-7s+3t\\
0&0&270&45-60t
\end{array}
\right].
$$
Back substitution now yields
$$
z=\frac16-\frac{2t}{9},\qquad
y=\frac{3-7s+3t+3z}{21}
=\frac16-\frac s3+\frac t9,
$$
and then
$$
\boxed{x=\frac16+\frac s3+\frac t9,\qquad
y=\frac16-\frac s3+\frac t9,\qquad
z=\frac16-\frac{2t}{9}.}
$$
All three pivots are nonzero. Thus \b[the solution exists uniquely for every $s,t$], the <matrix rank> is $3$, and the <kernel of a linear map> represented by $A$ is $\{\mathbf0\}$. The general <rank-nullity theorem> for a matrix with $m$ columns is
$$
\boxed{\operatorname{rank}A+\dim\ker A=m;}
$$
here $m=3$.
Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly when
$$
\boxed{x+y+z=\frac12.}
$$
When this holds, their unique values are $s=\tfrac32(x-y)$ and $t=\tfrac12(1+x+y-8z)$. \b[The allowable triples form an affine plane, not all of $\mathbb R^3$.]
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