Show that the four points `A(1,-1,1), B(2,3,1),C(1,2,3)` and `D(0,-2,3)` are coplanar. Find the equation of the plane containing them.

Explanation

We know that the equation of the plane passing through the ponts `A(x_(1),y_(1),z_(1)), B(x_(2),y_(2),z_(2))` and `C(x_(3),y_(3),z_(3))` is given by,
`|{:(x-x_(1),y-y_(1),z-z_(1)),(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),)(x_(3)-x_(1),y_(3)-y_(1),z_(3)-z_(1)):}|=0`.
Here,`(x_(1)=1, y_(1)=-1, z_(1)=1),(x_(2)=2,y_(2)=3,z_(2)=1)` and `(x_(3)=1,y_(3)=2,z_(3)=3)`.
`therefore` the equation of the plane passing through the points A, B and C is given by
`|{:(x-1,y+1,z-1),(2-1,3+1,1-1),(1-1,2+1,3-1)|=0 rArr |{:(x-1,y+1,z-1),(1,4,0),(0,3,2)|=0`
`rArr (x-1)(8-0)-(y+1)(2-0)+(z-1)(3-0)=0`
`rArr 8(x-1)-2(y+1)+3(z-1)=0 rArr 8x-2y+3z=13`.................(i)
Putting, `x=0, y=-2` and `z=3` in (i), we get
LHS `=(8xx0)-2 xx (-2)=3 xx 3=(0+4+9)=13`= RHS.
Thus, the point `(0,-2,3)` lies on the plane (i).
Hence, the given four points are coplanar and the the equation of the plane containing them is `8x=2y+3z=13`.