Given $\vec{F} = q(\vec{v} \times \vec{B})$. With $q=1$, $\vec{F} = \vec{v} \times \vec{B}$. Substituting vectors: $4\hat{i} - 20\hat{j} + 12\hat{k} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times (B\hat{i} + B\hat{j} + B_0\hat{k})$. Calculating the cross product using the determinant method: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\2 & 4 & 6 \\B & B & B_0 \end{vmatrix} = \hat{i}(4B_0 - 6B) - \hat{j}(2B_0 - 6B) + \hat{k}(2B - 4B)$. Comparing components with $4\hat{i} - 20\hat{j} + 12\hat{k}$, we get equations: $4B_0 - 6B = 4$, $-(2B_0 - 6B) = -20$, and $2B - 4B = 12$. Solving these yields $B = -6$ and $B_0 = -8$. Thus, $\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$.