## How many planes can be made to pass through three distict points if they are – a-collinear b-non-collinear what is the answer?

If the points are colinear then an infinite number of planes can be made to pass through them. If three distinct points are non-colinear then exactly one plane passes through them.

Let us look at the general case.

The general equation of a plane is ##ax+by+cz = d## for constants ##a##, ##b##, ##c##, ##d##.

For a given plane, this equation is unique only up to a constant factor. That is, the same plane is also described by the equation:

##kax+kby+kcz=kd## for any ##k != 0##

Let the three points be ##(x_1, y_1, z_1)##, ##(x_2, y_2, z_2)## and ##(x_3, y_3, z_3)##.

Then we have a system of three linear equations:

##ax_1 + by_1 + cz_1 = d##

##ax_2 + by_2 + cz_2 = d##

##ax_3 + by_3 + cz_3 = d##

In the general case, this is a little painful to deal with using substitution and/or elimination, so permit me to show how to do it using matrix arithmetic.

Let ##M = ((x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3))##

Then our three equations become:

##M((a), (b), (c)) = ((d), (d), (d))##

If we can find ##M^(-1)## then we get:

##((a), (b), (c)) = M^(-1)M((a), (b), (c)) = M^(-1)((d), (d), (d))##

Let:

##X_1 = det((y_2, z_2), (y_3, z_3)) = y_2z_3 – z_2y_3## ##Y_1 = z_2x_3 – z_3x_2## ##Z_1 = x_2y_3 – y_2x_3## ##X_2 = y_3z_1 – z_3y_1## ##Y_2 = z_3x_1 – x_3z_1## ##Z_2 = x_3y_1 – y_3x_1## ##X_3 = y_1z_2 – z_1y_2## ##Y_3 = z_1x_2 – x_1z_2## ##Z_3 = x_1y_2 – y_1x_2##

The determinant of ##M## is given by the formula:

##det(M) = x_1X_1 + y_1Y_1 + z_1Z_1##

If ##det(M) != 0## then ##M^-1## is given by:

##M^(-1) = 1/det(M)((X_1, X_2, X_3), (Y_1, Y_2, Y_3), (Z_1, Z_2, Z_3))##

How can this go wrong?

If the points are not distinct, then two of the rows of ##M## will be identical and ##det(M) = 0##

If the points are colinear, then ##det(M) = 0##

If the plane also passes through ##(0, 0, 0)## then ##det(M) = 0##

This last case is the one where ##d = 0##, so ##M^(-1)((d),(d),(d))## would be fairly useless anyway.