Determine whether $A (2, 2, 3)$, $B(4, 0, 7)$, $C (6, 3, 1)$ and $D (2, −3, 11)$ are in the same plane. 
a) Compute a suitable volume to determine whether $A  (2, 2, 3)$, $B 
(4, 0, 7)$, $C  (6, 3, 1)$ and $D  (2, −3, 11)$ are in the same plane.
b) Find the distance between the line $L$ through $A$, $B$ and the
  line $M$ through $C$, $D$.


My answer:
V = |(a-d) · ((b-d)x(c-d))| / 6
= -15/2
  Thoughts? Help?!?!

Alternatively: a) find the vectors:
$$\vec{AD}\{0,-5,8\}; \ \ \ \vec{BD}\{-2,-3,4\}; \ \ \ \vec{CD}\{-4,-6,10\}.$$
The volume of the tetrahedron $ABCD$:
$$V=\pm \frac16 \begin{vmatrix} 0 & -5 & 8 \\ -2 & -3 & 4 \\ -4 & -6 & 10 \end{vmatrix}=\pm \frac16 \cdot (-20)=\frac{10}{3}.$$
It implies the four points are not coplanar.
b) The equations of the lines $AB$ and $CD$:
$$\frac{x-2}{2}=\frac{y-2}{-2}=\frac{z-3}{4}; \ \ \ \frac{x-6}{-4}=\frac{y-3}{-6}=\frac{z-1}{10}.$$
Their parametric forms are:
$$x=2+2t, y=2-2t, z=3+4t; \\ x=6-4t', y=3-6t', z=1+10t'.$$
The radius-vectors $r_1,r_2$ of the points $A$ and $C$:
$$r_1=\{2,2,3\}; \ \ \ r_2=\{6,3,1\}$$
The directing-vectors $a_1, a_2$ of the lines $AB$ and $CD$:
$$a_1=\{2,-2,4\}; \ \ \ a_2=\{-4,-6,10\}.$$
Note the lines are not parallel, because $a_1\ne a_2$.
The formula of distance between non-parallel lines:
$$d=\frac{|(r_2-r_1)a_1a_2|}{\sqrt{(a_1 \times a_2)^2}}.$$
Calculate:
$$a_1 \times a_2=\{4, -36, -20\}; \\ (r_2-r_1)a_1a_2=4\cdot 4+1\cdot (-36)+(-2)\cdot (-20)=20.$$
At last:
$$d=\frac{20}{\sqrt{4^2+(-36)^2+(-20)^2}}=\frac{5}{\sqrt{107}}.$$
