Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-60 & 168 \\ -21 & 59}\] 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 \[A-3 I = \matrix{-60 & 168 \\ -21 & 59} - \matrix{3 & 0 \\0& 3 }=\matrix{ -63 & 168 \\ -21 & 56}\] This can be done as follows:
\[\begin{aligned}
\matrix{-63&168\\-21&56\\}&\sim\matrix{1&-{{8}\over{3}}\\-21&56\\}&{\blue{\begin{array}{c}-{{1}\over{63}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{8}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{-8\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(3\) equals \(\left\langle\cv{-8\\-3}\right\rangle\)