= Solution
If $Ax_p=b$, then $Ax=b$ is equivalent to $A(x-x_p)=0$. Thus the solution set is the <affine solution space of a linear equation> $\boxed{x_p+\ker A}$, and every vector of that form solves the equation.
For the <symmetric matrix> case, let $Au_i=\lambda_i u_i$. Symmetry gives
$$
\lambda_i(u_i\cdot u_j)=(Au_i)\cdot u_j=u_i\cdot(Au_j)=\lambda_j(u_i\cdot u_j).
$$
Distinct <eigenvalues> therefore imply $u_i\cdot u_j=0$. Normalizing the three vectors gives an <orthonormal basis>. Expand $x=\sum_i t_i u_i$ and $b=\sum_i b_i u_i$, with $b_i=b\cdot u_i$. The equation $(A-\lambda_kI)x=b$ becomes $(\lambda_i-\lambda_k)t_i=b_i$ for every $i$. Thus solvability is equivalent to $b_k=0$, and all solutions are
$$
\boxed{x=\sum_{i\ne k}\frac{b\cdot u_i}{\lambda_i-\lambda_k}u_i+\beta u_k,\qquad \beta\in\mathbb R.}
$$
This proves the claimed orthogonality condition and the full solution family rather than only a particular solution.
For the numerical <symmetric matrix>, direct multiplication gives $Au_1=0$ for $u_1=(1,1,1)^T/\sqrt3$, and $Au_2=-\sqrt2\,u_2$ for $u_2=(1,-1,0)^T/\sqrt2$. Their orthogonal complement is spanned by $u_3=(1,1,-2)^T/\sqrt6$, and multiplication gives $Au_3=\sqrt3\,u_3$. The supplied right side has $b\cdot u_1=0$, so the <linear system> is solvable. Its other coefficients are $b\cdot u_2=2$ and $b\cdot u_3=3\sqrt2$. Hence
$$
x=-\sqrt2\,u_2+\sqrt6\,u_3+\beta u_1
=(0,2,-2)^T+\frac\beta{\sqrt3}(1,1,1)^T.
$$
Equivalently, \b[all solutions are]
$$
\boxed{x=(0,2,-2)^T+t(1,1,1)^T,\qquad t\in\mathbb R.}
$$
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