Answer: (a) The coefficient matrix is \(A=\begin{pmatrix}a&3&1\\2&1&3\\-1&2&-5\end{pmatrix}\), so

\(\det(A)=a\begin{vmatrix}1&3\\2&-5\end{vmatrix}-3\begin{vmatrix}2&3\\-1&-5\end{vmatrix}+\begin{vmatrix}2&1\\-1&2\end{vmatrix}=-11a+26.\)

A unique solution exists when \(\det(A)\neq 0\). This would require \(-11a+26=0\), giving \(a=\tfrac{26}{11}\), which is not an integer. Since \(a\) is an integer, \(\det(A)\neq 0\), so the system has a unique solution.

Geometrically, the three planes intersect at exactly one point.

(b) \(a=4\).

(a) Write the system as \(A\mathbf{x}=\mathbf{b}\), where

\(A=\begin{pmatrix}a&3&1\\2&1&3\\-1&2&-5\end{pmatrix}.\)

A system of three linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.

So calculate

\(\det(A)=\begin{vmatrix}a&3&1\\2&1&3\\-1&2&-5\end{vmatrix}.\)

Expanding along the first row,

\(\det(A)=a\begin{vmatrix}1&3\\2&-5\end{vmatrix}-3\begin{vmatrix}2&3\\-1&-5\end{vmatrix}+\begin{vmatrix}2&1\\-1&2\end{vmatrix}.\)

Now,

\(\begin{vmatrix}1&3\\2&-5\end{vmatrix}=-5-6=-11,\quad \begin{vmatrix}2&3\\-1&-5\end{vmatrix}=-10+3=-7,\quad \begin{vmatrix}2&1\\-1&2\end{vmatrix}=4+1=5.\)

Hence

\(\det(A)=a(-11)-3(-7)+5=-11a+21+5=-11a+26.\)

For there not to be a unique solution, we would need

\(-11a+26=0\Rightarrow a=\tfrac{26}{11}.\)

But \(\tfrac{26}{11}\) is not an integer. Since \(a\) is given to be an integer, \(\det(A)\neq 0\) for every possible value of \(a\).

Therefore the system has a unique solution.

Geometrically, each equation represents a plane in three-dimensional space, so a unique solution means the three planes intersect in a single point.

(b) Substitute \(x=1\), \(y=4\), \(z=-2\) into the first equation:

\(a(1)+3(4)+(-2)=14.\)

So

\(a+12-2=14\Rightarrow a+10=14\Rightarrow a=4.\)

Thus the required value is \(a=4\).