The matrix
A=
\begin{pmatrix}
A11 & A12 & A13 \\
A21 & {\color{red}b} & A23\\
A31 & A32 & A33
\end{pmatrix}
has v =
\begin{pmatrix}
X \\ Y \\ Z
\end{pmatrix}
as an eigenvector.
{\color{red}b}.
\color{red} b
=
L2
For an eigenvector v there is an eigenvalue \color{orange}\lambda with
A \cdot v = {\color{orange}\lambda} \cdot v.
Thus, find {\color{red}b}, such that this equation is satisfied, i.e. there is
{\color{orange}\lambda} with
\begin{pmatrix}
A11 & A12 & A13 \\
A21 & {\color{red}b} & A23\\
A31 & A32 & A33
\end{pmatrix} \cdot \begin{pmatrix}
X \\ Y \\ Z
\end{pmatrix} = {\color{orange}\lambda} \cdot \begin{pmatrix}
X \\ Y \\ Z
\end{pmatrix}.
Calculating the left-hand side gives
\begin{pmatrix}
A11 & A12 & A13 \\
A21 & {\color{red}b} & A23\\
A31 & A32 & A33
\end{pmatrix} \cdot \begin{pmatrix}
X \\ Y \\ Z
\end{pmatrix} =
\begin{pmatrix}
negParens(A11) \cdot negParens(X) +
negParens(A12) \cdot negParens(Y) +
negParens(A13) \cdot negParens(Z) \\
negParens(A21) \cdot negParens(X) +
{\color{red}b} \cdot negParens(Y) +
negParens(A23) \cdot negParens(Z) \\
negParens(A31) \cdot negParens(X) +
negParens(A32) \cdot negParens(Y) +
negParens(A33) \cdot negParens(Z)
\end{pmatrix} =
\begin{pmatrix}
A11 * X + A12 * Y + A13 * Z \\
{\color{red}b} \\ 0
\end{pmatrix}= {\color{orange}L2} \cdot
\begin{pmatrix}
X \\
\dfrac{\color{red}b}{L2} \\ 0
\end{pmatrix}.
With {\color{red}b} = {\color{orange}\lambda} = {\color{orange}L2}
the above equation A \cdot v = {\color{orange}\lambda} \cdot v holds.
PS: The eigenvalues of a (upper or lower) triangle matrix are the entries on the diagonal, i.e.
here we already knew that {\color{orange}\lambda} \in \{A11,
{\color{red}b}, A33\}.