If $\beta = \{u_1,u_2,…,u_n\}$ be an ordered basis for a finite-dimensional vector space $V$ and for $x \in V,$ let $a_1,a_2,…,a_n$ be the unique scalars such that $$x = \sum_{i=1}^{n} a_iu_i$$
then coordinate vector of $x$ relative to $\beta$ is defined as: $$[x]_\beta = \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}$$
So, here considering $e= \{e_1,e_2,e_3\}$ as an ordered basis for vector space $V$ with $e_1=1,e_2=t-1,e_3=(t-1)^2$
Now, $v \in V$ is given as $v=2t^2-5t+6.$ So, we can write $v$ as $\sum_{i=1}^{3} a_ie_i.$ So,
$2t^2-5t+6 = a_1e_1 +a_2e_2+a_3e_3$
$2t^2-5t+6 = a_1\times 1 +a_2(t-1)+a_3(t-1)^2$
$2t^2-5t+6 = a_1 + a_2(t-1) + a_3 (t^2+1-2t)$
$2t^2-5t+6 = a_3t^2+(a_2-2a_3)t+(a_1-a_2+a_3)$
On comparing the coefficients both sides,
$a_3=2; a_2=-1$ and $a_1=3$
Hence, $$[v]_e = \begin{pmatrix} a_1\\ a_2\\a_3 \end{pmatrix} = \begin{pmatrix} 3\\ -1\\2 \end{pmatrix}$$
