# Show that the plane that passes through the three points A=(a1,a2,a3) B=(b1,b2,b3) C=(c1,c2,c3) consists of the points P=(x,y,z) given by |[a_1-x,a_2-y, a_3-z],[b_1-x,b_2-y, b_3-z], [c_1-x,c_2-y, c_3-z]|=0

Show that the plane that passes through the three points
A=(a1,a2,a3)
B=(b1,b2,b3)
C=(c1,c2,c3)
consists of the points P=(x,y,z) given by
$|\begin{array}{ccc}{a}_{1}-x& {a}_{2}-y& {a}_{3}-z\\ {b}_{1}-x& {b}_{2}-y& {b}_{3}-z\\ {c}_{1}-x& {c}_{2}-y& {c}_{3}-z\end{array}|$
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Example:
(a1-x)[(b2-y)(c3-z)-(c2-y)(b3-z)]=(a1-x)[b2c3-(b2z+c3y)+yz-(c2b3-c2z-b3y+yz)]=
(a1-x)[b2c3-c2b3-(b2-c2)z-(c3-b3)y]. The higher order productsin xyz, xy, yz, xz cancel out and only linear terms remain. Thedeterminate has the form of a point - normal form of the planeequation. E+Fx+Gy+Hz=0.
D has the geometrical meaning of the distance of A along thenormal form of the cross product of B with C as the normal of theplane: A (dot) (BxC).
F=a2b1-a1b2-a2c1+b2c1+a1c2-b1c2.G=-a3b1+a1b3+a3c1-b3c1-a1c3+c1c3.H=a3b2-a2b3-a3c2+b3c2+a2c3-b2c3.
(F,G,H) is the normal of the plane containing the three pointsA,B,C.
The determinate can be interpreted intermediate as theparallelepiped product: (A-x) (dot) (B-x)X(C-x). So A is in theplane and BXC is the normal made two vectors in the plane. With xand A,B,C all vectors in the plane ful fill the condition. With x inthe plane the difference vectors are also in the plane, and thecross product forms a normal to the plane.