To find the positive value of \(a\) for which the given system of linear homogeneous equations has a non-trivial solution, we start by using the condition for non-trivial solutions: the determinant of the coefficient matrix must be zero.
The system of equations given is:
\[x + ay + z = 0, \\ ax + 2y - z = 0, \\ 2x + 3y + z = 0\]The coefficient matrix \(A\) of the system is:
\[A = \begin{pmatrix} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{pmatrix}\]The determinant of \(A\) is given by:
\[\text{det}(A) = \begin{vmatrix} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix}\]Expanding this determinant along the first row, we get:
\[\text{det}(A) = 1 \cdot \left|\begin{array}{cc} 2 & -1 \\ 3 & 1 \end{array}\right| - a \cdot \left|\begin{array}{cc} a & -1 \\ 2 & 1 \end{array}\right| + 1 \cdot \left|\begin{array}{cc} a & 2 \\ 2 & 3 \end{array}\right| \]