# Consider the points A(-5,1,3), B(4,1,2), C(1,-2,1) and D(9,0,1). Show that these points are coplanar by using vector products.

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The points given are A(-5,1,3), B(4,1,2), C(1,-2,1) and D(9,0,1).

The vector AB is (4+5)i + (1-1)j + (2-3)k = 9i - k

AC = (1+5)i + (-2-1)j + (1 - 3)k = 6i - 3j - 2k

The cross product `ABoxAC` gives the normal and is:` [[i, j, k],[9, 0, -1],[6, -3, -2]]`

= i(0 - 3)-j(-18+6) + k(-27)

= -3i + 12j - 27k

This gives the equation of the plane through A, B and C as:

-3(x + 5) + 12(y - 1) -27(z - 3) = 0

=> -3x - 15 + 12y - 12 - 27z + 81 = 0

=> -3x + 12y - 27z + 54 = 0

Substituting the coordinates of the point D, -3*9 + 0 - 27 + 54 = 0

This shows that the point D lies on the plane passing through A, B and C.

**The 4 given points are co-planar.**