Answer: The system has a unique solution for all values of \(a\) except \(a=8\) and \(a=18\).

For \(a=18\), the system has no solution.

For \(a=8\), the system has infinitely many solutions, and the one with \(x+y+z=1\) is \(x=\frac{1}{3},\ y=\frac{5}{12},\ z=\frac{1}{4}\).

Consider the coefficient matrix \(A\) of the system:

\(A=\begin{pmatrix}1&4&12\\2&a&12\\3&12&2a\end{pmatrix}.\)

A unique solution exists exactly when \(\det A \neq 0\).

Expand the determinant by row operations:

Subtract \(2\) times the first row from the second, and \(3\) times the first row from the third:

\(\begin{pmatrix}1&4&12\\0&a-8&-12\\0&0&2a-36\end{pmatrix}.\)

So

\(\det A = 1\cdot (a-8)(2a-36).\)

Hence \(\det A \neq 0\) when \(a \neq 8\) and \(a \neq 18\). Therefore the system has a unique solution for all values of \(a\) except \(8\) and \(18\).

Now take \(a=18\). The equations become

\(x+4y+12z=5,\)

\(2x+18y+12z=17,\)

\(3x+12y+36z=10.\)

From the row-reduced form above, the third diagonal entry is \(2a-36=0\), and the corresponding augmented equation is inconsistent. Equivalently, eliminating \(x\) and \(y\) gives a contradiction of the form \(0=\text{nonzero}\). So there is no solution when \(a=18\).

Now take \(a=8\). The system becomes

\(x+4y+12z=5,\)

\(2x+8y+12z=7,\)

\(3x+12y+16z=10.\)

Eliminate using the first equation:

Second minus twice the first gives \(-12z=-3\), so \(z=\frac14\).

Substituting into the first equation gives

\(x+4y+3=5\), so \(x+4y=2\).

Thus the solutions are

\(x=\lambda,\quad y=\frac{2-\lambda}{4},\quad z=\frac14,\)

for any real \(\lambda\). Since \(\lambda\) is free, there are infinitely many solutions.

Now impose \(x+y+z=1\):

\(\lambda+\frac{2-\lambda}{4}+\frac14=1.\)

Multiply by \(4\):

\(4\lambda+2-\lambda+1=4\), so \(3\lambda=1\), hence \(\lambda=\frac13\).

Therefore

\(x=\frac13,\quad y=\frac{5}{12},\quad z=\frac14.\)