Question

Find $x$ such that the four points $A(3,2,1),B(4,x,5),C(4,2,-2)$ and $D(6,5,-1)$ are coplanar.

Answer 1

Let
$\overrightarrow{OA}=3\hat{i}+2\hat{j}+\hat{k};\overrightarrow{OB}=4\hat{i}+x\hat{j}+5\hat{k}$

$\overrightarrow{OC}=4\hat{i}+2\hat{j}-2\hat{k};\overrightarrow{OD}=6\hat{i}+5\hat{j}-\hat{k}$

$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}+(x-2)\hat{j}+4\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\hat{i}+0\hat{j}-3\hat{k}$

$\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=3\hat{i}+3\hat{j}-2\hat{k}$

To find $x$, given vectors are coplanar

$=\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=0$

$\left|\begin{array}{ccc}1& x-2& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=0$

Expanding the determinant
$\Rightarrow 1(0+9)-(x-2)(-2+9)+4(3-0)=0$

$\Rightarrow 9-(x-2)\left(7\right)+12=0$

or $-7x+14+21=0$

$\Rightarrow -7x=-35$

or $x=5$.