Let $0 \neq a_1,a_2,a_3 \in \mathbb{R}^3$ such that every pair of vectors are not collinear and
$3a_1+2a_2+a_3=0$
Find the kernel of the matrix $A=\begin{bmatrix} a_1& a_2& a_3\end{bmatrix}$ whose $i$-th column is the vector $a_i$.
Since two of vectors are linear independent and $dim\ ker(A)>1$ then $dim\ ran(A)=2$ and $dim\ ker(A)=1$, so that $ker(A)=span(3,2,1)$, is this ok?
As the columns of the matrix are linearly dependent (because $3a_1 + 2 a_2 + a_3=0$), $dim\ ker(A) \geq 1$. (You wrote $>$ and it should be $\geq$.)
Suppose $dim\ ker(A) > 1$, then $dim\ ran(A) < 3 - 1 = 2$, so $dim\ ran(A) \leq 1$. As $A$ is not the zero matrix, $dim\ ran(A) =1$, but that implies, for example, that $a_1$ and $a_2$ are collinear, which is absurd.
So we conclude that $dim\ ker(A)=1$. As $(3\ 2\ 1) \in ker(A)$, we conclude that $span(3\ 2\ 1) = ker(A)$.