Answer: The system has a unique solution for all values of \(a\) except \(a=-3\) and \(a=5\).

When \(a=-3\), the system has no solution.

When \(a=5\):

(i) the number of solutions is infinite;

(ii) the solution satisfying \(x+y+z=2\) is \(x=-1\), \(y=\frac{7}{2}\), \(z=-\frac{1}{2}\).

Write the augmented matrix for the system:

\(\left(\begin{array}{cccc}1 & -2 & -2 & -7 \\ 2 & a-9 & -10 & -11 \\ 3 & -6 & 2a & -29\end{array}\right)\).

Use the first row to eliminate the \(x\)-terms from the second and third rows:

\(R_2 \leftarrow R_2-2R_1\), so

\((2,\ a-9,\ -10,\ -11) - 2(1,\ -2,\ -2,\ -7) = (0,\ a-5,\ -6,\ 3)\).

Also \(R_3 \leftarrow R_3-3R_1\), so

\((3,\ -6,\ 2a,\ -29) - 3(1,\ -2,\ -2,\ -7) = (0,\ 0,\ 2a+6,\ -8)\).

Hence the matrix is row-equivalent to

\(\left(\begin{array}{cccc}1 & -2 & -2 & -7 \\ 0 & a-5 & -6 & 3 \\ 0 & 0 & 2a+6 & -8\end{array}\right)\).

A unique solution exists when there is a pivot in every variable column, so we need \(a-5 \neq 0\) and \(2a+6 \neq 0\). Therefore \(a \neq 5\) and \(a \neq -3\).

So the system has a unique solution for all real \(a\) except \(a=-3\) and \(a=5\).

For \(a=-3\), the last row becomes \(0z=-8\), which is impossible. Therefore the system has no solution.

For \(a=5\), the second row becomes \(0\cdot y-6z=3\), so \(z=-\frac{1}{2}\). The first two non-zero equations reduce to

\(x-2y-2z=-7\) and \(x-2y=-8\).

Since \(z=-\frac{1}{2}\), the third equation is consistent with the first two, so there are infinitely many solutions.

To find the solution with \(x+y+z=2\), use \(z=-\frac{1}{2}\), giving

\(x+y=2-\left(-\frac{1}{2}\right)=\frac{5}{2}\).

Also from \(x-2y=-8\), subtracting gives

\((x+y)-(x-2y)=\frac{5}{2}-(-8)\), so \(3y=\frac{21}{2}\), hence \(y=\frac{7}{2}\).

Then \(x=\frac{5}{2}-\frac{7}{2}=-1\).

So the required solution is \(x=-1\), \(y=\frac{7}{2}\), \(z=-\frac{1}{2}\).