= Solution
Let $e_1,\ldots,e_n$ be the standard <basis> and define $a_j=f(e_j)$. By linearity,
$$
f(x)=f\!\left(\sum_{j=1}^n x_je_j\right)=\sum_{j=1}^n x_jf(e_j)=a\cdot x.
$$
Thus every <linear functional> has this form. Its representing vector is unique because evaluating $a\cdot x=b\cdot x$ at each $e_j$ gives $a_j=b_j$.
For the prescribed <kernel of a linear map>, the third listed vector is the first minus the second, so it adds no independent constraint. The first two require
$$
a_1+a_2+a_3-a_4=0,\qquad2a_1-a_2-2a_4=0.
$$
Taking $a_1=s$ and $a_4=t$ gives $a_2=2s-2t$, $a_3=-3s+3t$. Consequently \b[all possible representing vectors] are
$$
\boxed{a=s(1,2,-3,0)^T+t(0,-2,3,1)^T,\qquad s,t\in\mathbb R.}
$$
Conversely, every such vector is perpendicular to each of the three prescribed vectors, so its <linear functional> vanishes on them. This also includes the zero <linear functional>.
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