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.