Question

Find $x$ such that the four points $A\left(321\right),B\left(4x5\right),C(42-2),D(66-1)$ are coplanar.

Answer 1

$\overline{AB}=\overline{OB}-\overline{OA}=(1x-24)$
$\overline{AB}=\overline{OC}-\overline{OA}=(10-3)$
$\overline{AD}=\overline{OD}-\overline{OA}=(33-2)$
Given $\left[\overline{AB}\overline{AC}\overline{AD}\right]=0$
$\left|\begin{array}{ccc}1& x-2& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=0$
$1(0+9)-(x-2)(-2+9)+4(0-3)=0$
$9+(2-x)\left(7\right)-12=0$
$7x=35\Rightarrow x=5$