Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{27 & -56 \\ 12 & -25}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), that is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\).
We can do this through row reduction of the matrix \[A-3 I = \matrix{27 & -56 \\ 12 & -25} - \matrix{3 & 0 \\0& 3 }=\matrix{ 24 & -56 \\ 12 & -28}\] This can be done as follows:
\[\begin{aligned}
\matrix{24&-56\\12&-28\\}&\sim\matrix{1&-{{7}\over{3}}\\12&-28\\}&{\blue{\begin{array}{c}{{1}\over{24}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{7}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{7\\3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(3\) equals \(\left\langle\cv{7\\3}\right\rangle\)