On the method described by Purcell for finding the magnetic field by measuring the force on a test particle

The linear system you've written, $$ \vec f= \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} =q \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix} =q(\vec v\times) \vec B $$ is indeterminate, and it does not have a unique solution. You can easily verify this by noticing that the determinant is zero, since the matrix $M=(\vec v\times)$ is antisymmetric, but the transpose can't change the determinant, so $$ \det(M) = \det(M^T) = \det(-M) = (-1)^3\det(M) = -\det(M), $$ which requires $\det(M)=0$.

Alternatively, you can just notice that it is unable to distinguish a magnetic field parallel to the velocity from a vanishing field (since both give zero force). This is seen most simply by choosing your coordinate axes so that $\vec v$ lies along the $x$ axis, so that the system of equations reads $$ \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} = qv_1 \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix}, $$ which is clearly sufficient to figure out $B_2$ and $B_3$ from $f_2$ and $f_3$, but cannot say anything about $B_1$.

The system of equations

$$f_{1}=q(v_{2}B_{3}-v_{3}B_{2})$$ $$f_{2}=q(v_{1}B_{3}-v_{3}B_{1})$$ $$f_{3}=q(v_{1}B_{2}-v_{2}B_{1})$$

Can be rewritten as $$\mathbf{f}=q\mathbf{M} \mathbf{B}$$ where $$\mathbf{M}=\left( \begin{array}{ccc} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \\ \end{array} \right)$$

Note that $\det(\mathbf{M})=0$ so the system is not linearly independent. So even though you do have three equations in three unknowns there is not a unique solution. You need at least two non-zero $\mathbf{v}$ to solve it.