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Coplanarity of four points.

The points $A(x_1 ,y_1,z_1), B(x_2,y_2,z_2),C(x_3,y_3,z_3),D(x_4,y_4,z_4) $ are coplanar then

how can we write it like this .

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    What's the question? – 2017-01-20
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    Are you asking why does the determinant satisfy the condition? – 2017-01-20
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    @Rohan yes my doubt is that – 2017-01-20
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    @AlessandroCodenotti why does the determinant satisfy the condtion – 2017-01-20
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    The absolute value of the determinant is the volume of the solid defined by AB, AC and AD. So when it is $0$ the solid is flat and the forum points are coplanar – 2017-01-20

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The volume of the tetrahedron formed by four points $P (x_1, y_1, z_1); Q (x_2, y_2, z_2); R (x_3, y_3, z_3); S (x_4, y_4, z_4) $ is given by $$6\Delta = \begin {vmatrix} x_1 & y_1 & z_1 & 1\\ x_2 & y_2 & z_2 & 1\\ x_3 & y_3 & z_3 & 1\\ x_4 & y_4 & z_4 & 1\\ \end {vmatrix}$$ $$ = \begin {vmatrix} x_1 & y_1 & z_1 & 1\\ x_2-x_1 & y_2-y_1 & z_2-z_1 & 0\\ x_3-x_1 & y_3-y_1 & z_3-z_1 & 0\\ x_4-x_1 & y_4-y_1 & z_4-z_1 & 0\\ \end {vmatrix} $$

If the points are coplanar then the volume of the tetrahedron formed is zero, i.e, $\Delta =0$ the condition automatically follows. Hope it helps.

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    That $\Delta$ should be $6\Delta$. – 2017-01-20
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Vectors $\overrightarrow{AB} , \overrightarrow{AC}, \overrightarrow {AD}$ are coplanar if matrix $M= \begin{bmatrix} \overrightarrow{AB} & \overrightarrow{AC} & \overrightarrow {AD} \end{bmatrix}$ consists of columns linearly dependent i.e $\det(M)=0$.

Of course if $\det(M)=0$ then also $\det(M^T)=0$.