Equations of a plane

Here is one way
Let the general equation of the plane be
[tex] Ax + By + Cz + d =0 [/tex]
Now the plane passes through 3 points. So substitute the points in the above equation. You will have three equations. Solve for A,B and C in terms of d. Substitute them back and you will get your required equation

 

Not really, perhaps your book prefers to use the same vectors every time but as you say, you can do it several ways and they should all give the same.

Another way to get the carthesian equation using lineair algebra would be expanding the following determinant to the first row.

[tex]\left| {\begin{array}{*{20}c}
x & y & z & 1 \\
{x_1 } & {y_1 } & {z_1 } & 1 \\
{x_2 } & {y_2 } & {z_2 } & 1 \\
{x_3 } & {y_3 } & {z_3 } & 1 \\
\end{array}} \right| = 0[/tex]

In this determinant, you can also replace points by vectors giving the direction of the plane, the 1 in the last column then becomes a 0. (This is easy to see since the difference of 2 points gives such a vector)

 

In any case, you shouldn't "check your answers" by comparing them to the books answers- If your equation is correct, then the coordinates of the three points will all satisfy the equation!