Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-51 & 112 \\ -24 & 53}\] 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 \[A+3 I = \matrix{-51 & 112 \\ -24 & 53} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -48 & 112 \\ -24 & 56}\] This can be done as follows:
\[\begin{aligned}
\matrix{-48&112\\-24&56\\}&\sim\matrix{1&-{{7}\over{3}}\\-24&56\\}&{\blue{\begin{array}{c}-{{1}\over{48}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{7}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+24R_1\end{array}}} \end{aligned}\] 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 \(\left\langle\cv{-7\\-3}\right\rangle\)