The angular momentum $\vec{L}$ of a particle is given by the cross product of its position vector $\vec{r}$ and linear momentum vector $\vec{P}$. $\vec{L} = \vec{r} \times \vec{P}$ $\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\1 & 2 & -1 \\3 & 4 & -2 \end{vmatrix}$ Evaluating the determinant: $\vec{L} = \hat{i}[(2)(-2) - (-1)(4)] - \hat{j}[(1)(-2) - (-1)(3)] + \hat{k}[(1)(4) - (2)(3)]$ $\vec{L} = \hat{i}(-4 + 4) - \hat{j}(-2 + 3) + \hat{k}(4 - 6)$ $\vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k}$ Since the x-component of the angular momentum is zero ($L_x = 0$), the vector $\vec{L}$ lies entirely in the Y-Z plane. Any vector lying in the Y-Z plane is perpendicular to the X-axis.