Answer: \(a=-1\).

(i) When \(b=-5\), the general solution is

(ii) When \(b\ne -5\), the system is inconsistent. Geometrically, the three planes intersect pairwise in parallel lines and form a prism.

For the system to have no unique solution, the coefficient matrix must be singular, so its determinant is zero:

\(\det\begin{pmatrix}1 & -1 & 2\\ 1 & a & -3\\ 1 & -1 & 7\end{pmatrix}=0\).

Subtract the first row from the second and third rows:

\(\det\begin{pmatrix}1 & -1 & 2\\ 0 & a+1 & -5\\ 0 & 0 & 5\end{pmatrix}=0\).

Hence

\(5(a+1)=0\),

so

\(a=-1\).

Now take \(a=-1\). Then the system is

\(x-y+2z=4\),

\(x-y-3z=b\),

\(x-y+7z=13\).

Subtract the first equation from the second and third:

\(-5z=b-4\),

\(5z=9\).

When \(b=-5\):

\(-5z=-9\), so \(z=\frac95\).

Substitute into \(x-y+2z=4\):

\(x-y=4-\frac{18}{5}=\frac25\).

Let \(y=t\). Then \(x=t+\frac25\), and so

\((x,y,z)=\left(\frac25,0,\frac95\right)+t(1,1,0),\quad t\in\mathbb R.\)

When \(b\neq -5\):

The two reduced equations for \(z\) are inconsistent, so the system has no solution.

Geometrically, the three planes intersect pairwise in parallel lines, so they form a prism.