Answer: The system has a unique solution when the coefficient matrix is invertible, so its determinant must be non-zero.

For the equations
\(ax+y+2z=0\), \(3x-2y=4\), and \(3x-4y-6az=14\), the coefficient matrix is

\(\begin{pmatrix} a & 1 & 2 \\ 3 & -2 & 0 \\ 3 & -4 & -6a \end{pmatrix}\).

Its determinant is

\(\begin{aligned} \det &= a\begin{vmatrix}-2 & 0 \\ -4 & -6a\end{vmatrix}-1\begin{vmatrix}3 & 0 \\ 3 & -6a\end{vmatrix}+2\begin{vmatrix}3 & -2 \\ 3 & -4\end{vmatrix} \\ &= a\big(( -2)(-6a)-0\cdot(-4)\big)-\big(3(-6a)-0\cdot 3\big)+2\big(3(-4)-(-2)(3)\big) \\ &= a(12a)-(-18a)+2(-12+6) \\ &= 12a^2+18a-12. \end{aligned}\)

For a unique solution,

\(12a^2+18a-12 \neq 0\).

Divide by 6:

\(2a^2+3a-2 \neq 0\).

Factorise:

\((2a-1)(a+2) \neq 0\).

So the system has a unique solution when \(a \neq \frac12\) and \(a \neq -2\).

Let the coefficient matrix be \(A\). A unique solution exists exactly when \(\det(A)\neq 0\).

Here

\(A=\begin{pmatrix} a & 1 & 2 \\ 3 & -2 & 0 \\ 3 & -4 & -6a \end{pmatrix}.\)

Compute the determinant by expanding along the first row:

\(\begin{aligned} \det(A) &= a\begin{vmatrix}-2 & 0 \\ -4 & -6a\end{vmatrix} -1\begin{vmatrix}3 & 0 \\ 3 & -6a\end{vmatrix} +2\begin{vmatrix}3 & -2 \\ 3 & -4\end{vmatrix} \\ &= a\big((-2)(-6a)-0\cdot(-4)\big) -\big(3(-6a)-0\cdot 3\big) +2\big(3(-4)-(-2)(3)\big) \\ &= 12a^2+18a-12. \end{aligned}\)

Hence we need

\(12a^2+18a-12\neq 0.\)

Dividing by 6 gives

\(2a^2+3a-2\neq 0,\)

and factorising:

\(\begin{aligned} 2a^2+3a-2 &= (2a-1)(a+2). \end{aligned}\)

So the determinant is non-zero when

\((2a-1)(a+2)\neq 0,\)

which means

\(a\neq \frac12\) and \(a\neq -2\).