Step-by-step Solution:

\[ \text{Let the given points be } A(1, -1, 0), B(2, 1, -2), C(-1, 1, 2) \] \[ \text{Find two direction vectors:} \] \[ \vec{AB} = B - A = (2 - 1, 1 - (-1), -2 - 0) = (1, 2, -2) \] \[ \vec{AC} = C - A = (-1 - 1, 1 - (-1), 2 - 0) = (-2, 2, 2) \] \[ \text{Find the normal vector } \vec{N} \text{ using the cross product:} \] \[ \vec{N} = \vec{AB} \times \vec{AC} \] \[ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ -2 & 2 & 2 \end{vmatrix} \] \[ \text{Expanding along the first row:} \] \[ \vec{N} = \hat{i} \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ -2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -2 & 2 \end{vmatrix} \] \[ \text{Compute the determinants:} \] \[ \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} = (2)(2) - (-2)(2) = 4 + 4 = 8 \] \[ \begin{vmatrix} 1 & -2 \\ -2 & 2 \end{vmatrix} = (1)(2) - (-2)(-2) = 2 - 4 = -2 \] \[ \begin{vmatrix} 1 & 2 \\ -2 & 2 \end{vmatrix} = (1)(2) - (2)(-2) = 2 + 4 = 6 \] \[ \text{Thus,} \] \[ \vec{N} = 8\hat{i} + 2\hat{j} + 6\hat{k} \] \[ \vec{N} = (8, 2, 6) \] \[ \text{Final Answer: The required perpendicular vector to the plane is } (8, 2, 6) \]