Find the area of the triangle with vertices $(0,7,10),$ $(-1,6,6),$ and $(-4,9,6).$
The area of a triangle with vertices $(x_1,y_1,z_1),$ $(x_2,y_2,z_2),$ and $(x_3,y_3,z_3)$ is $\frac{1}{2}$ the magnitude of the cross product of the vectors $\begin{pmatrix}x_2-x_1 \\ y_2-y_1 \\ z_2-z_1\end{pmatrix}$ and $\begin{pmatrix}x_3-x_1 \\ y_3-y_1 \\ z_3-z_1\end{pmatrix}$. So let's first compute the vectors.  From $(0,7,10)$ to $(-1,6,6)$, the difference is $\begin{pmatrix}-1 \\ -1 \\ -4\end{pmatrix}$.  From $(0,7,10)$ to $(-4,9,6)$, the difference is $\begin{pmatrix}-4 \\ 2 \\ -4\end{pmatrix}$. So now we need to compute the cross product of those two vectors.  That is, we need to compute $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & -4 \\ -4 & 2 & -4 \end{vmatrix}$. So the $\mathbf{i}$ entry is $4-(-8)$, which is $12$. So the cross product is $\begin{pmatrix}12 \\12 \\ -6\end{pmatrix}$. So the magnitude of the cross product is $\sqrt{12^2+12^2+(-6)^2}$, which is $\sqrt{144+144+36}$, which is $\sqrt{324}$, which is 18. So half of that is 9. So the area of the triangle is 9. Right, so we have the final answer.

