# For what value of k do the four points (1, 1, -1), (0, 3, -2), (-2, 1, 0), and (k, 0, 2) all lie in a plane?

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You should consider the condition for the 4 points to be coplanar,hence, you need to evaluate the volume of tetrahedron defined by the points and then you need to solve the equation V=0, such that:

`V = [(1, 1, -1, 1),(0, 3, -2, 1), (-2, 1, 0, 1), (k, 0, 2, 1)]`

`V = (-1)^(1+1)*1*[(3,-2,1),(1,0,1),(0,2,1)] + (-1)^(3+1)*(-2)*[(1,-1,1),(3,-2,1),(0,2,1)] + (-1)^(4+1)*k*[(1,-1,1),(3,-2,1),(1,0,1)]`

`V = (2-6+2) - 2(-2+6-2+3) - k(-2-1+2+3)`

`V = -2 - 10 - 2k`

Hence, all four points are coplanar if the volume of tetrahedron is zero such that:

`-2 - 10 - 2k = 0 => -12- 2k = 0 => -2k = 12 => k = -6`

**Hence, evaluating the value of k for the given points to be coplanar yields `k = -6` .**

Could my first question be solved by calculating on when the volume of a parallelepiped or a cube has V=0 aswell? And how would the formula of V be then?

Why is the volume of the tetrahedron the augumented matrix of the four points with 1's in last column? The other steps I understand.