In the product
$$\vec{F} = q(\vec{v} \times \vec{B}) = q\vec{v} \times \left(B\hat{i} + B\hat{j} + B_0\hat{k}\right)$$
For $q = 1$ and $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$
What will be the complete expression for $\vec{B}$ ?
Answer: (C) $-6\hat{i} - 6\hat{j} - 8\hat{k}$
$$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ B & B & B_0 \end{vmatrix} = (4B_0 - 6B)\hat{i} + (6B - 2B_0)\hat{j} + (2B - 4B)\hat{k}$$
Compare with $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$:
$\hat{k}$: $-2B = 12$, so $B = -6$.
$\hat{i}$: $4B_0 + 36 = 4$, so $B_0 = -8$.
Check $\hat{j}$: $6(-6) - 2(-8) = -36 + 16 = -20$. ✓
$$\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$$
Solution by Sreeraj P, M.Sc Physics