For which \color{red} a does the linear system Ax = 0
with
A= \begin{pmatrix} {\color{red} a} & B & C \\
R & D & E \\
S & F & G
\end{pmatrix}
have a nontrivial solution x = \begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} \neq \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.
{\color{red}a} =
A
There's a nontrivial solution x = \begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} \neq \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} if and only if the determinant of A is not zero.
For a 3 \times 3 - matrix (and only for those) the determinant can be computed with the rule of Sarrus .
This yields \det = {\color{red}a}\cdot negParens(D*G) + B*E*S + C*R*F
- C*D*S - B*R*G - {\color{red}a}\cdot negParens(E*F).
Solving the equation \det = 0 gives
{\color{red}a} = A.