I'm having trouble understanding why i'm getting a different result than the book for the given problem.
Determine a 4d vector $\langle N, D\rangle$ corresponding to the plane that passes through the three points $A = \langle 1, 2, 0 \rangle$, B = $\langle 2, 0, -1 \rangle$, and $C = \langle 3, -2, -1 \rangle$
The answer they have indicates that it is $\langle 2, 1, 0, -4 \rangle$
The steps i've take are as follows: $$\begin{align} \vec{AB} & = \langle B_x - A_x, B_y - A_y, B_z - A_z \rangle \\ & = \langle 2 - 1, 0 - 2, (-1) - 0 \rangle \\ & = \langle 1, - 2, -1 \rangle \\ \vec{AC} & = \langle C_x - A_x, C_y - A_y, C_z - A_z \rangle \\ & = \langle 3 - 1, (-2) - 2, (-1) - 0 \rangle \\ & = \langle 2, - 4, -1 \rangle \end{align}$$
Identify the Normal Vector (N) $$ \begin{align} N & = \vec{AB} \times \vec{AC} \\ & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 2 & -4 & -1 \end{vmatrix} \\ & = \hat{i}(((-2)\cdot(-1)) - ((-1)\cdot(-4))) + \hat{j}((-1\cdot2) - (1\cdot(-1))) + \hat{k}((1\cdot(-4)) - ((-2)\cdot2)) \\ & = \hat{i}(-2) + \hat{j}(-1) + \hat{k}(0) \\ & = \langle-2, -1, 0\rangle \end{align}$$
Identify the perpendicular distance (D) $$\begin{align} D & = -N\cdot A \\ & = -\begin{bmatrix}-2, -1, 0\end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \\ & = 2\cdot1 + 1\cdot2 + 0 \cdot 0 \\ & = 4 \end{align}$$
Which results in $ \langle N, D\rangle = \langle -2, -1, 0, 4 \rangle$.
Remember that a non-zero scalar multiple of a homogeneous coordinate vector is equivalent to it. The answer you came up with only differs from the key’s in its sign, that is, it differs by a factor of $-1$, so your answer matches. In effect, what you’re doing is finding the coefficients of an equation of the form $ax+by+cz+d=0$ that all three points satisfy. Multiplying both sides of this equation by a non-zero constant doesn’t affect its solution set.
If you’re familiar with row-reduction, a.k.a., Gaussian elimination, then you can find the vector that represents the plane directly by assembling the homogeneous 4-vectors of the points as the rows of a matrix and computing its null space. Specifically, $$\begin{bmatrix}1&2&0&1\\2&0&-1&1\\3&-2&-1&1\end{bmatrix} \to \begin{bmatrix}1&0&0&\frac12\\0&1&0&\frac14\\0&0&1&0\end{bmatrix}$$ from which you can read that the homogeneous null vector is $\left[\frac12:\frac14:0:-1\right] = [2:1:0:-4]$.