equation of a plane in space though 3 points We have the points $M(1,2,3),N(2,1,5), P(4,3,2)$ is it a correct way to find the equation of the plane using the determinant $\Delta$
$$\Delta=\begin{array}{|cccc|}
x & y & z & 1\\
1& 2 & 3 & 1\\
2 & 1 & 5 & 1\\
4 & 3 & 2 & 1
\end{array}=0$$
If so, what is the intuition behind this method of finding equation of a plane?
Thanks in advance!
 A: $$\Delta=\begin{array}{|cccc|}
x & y & z & 1\\
1& 2 & 3 & 1\\
2 & 1 & 5 & 1\\
4 & 3 & 2 & 1
\end{array}=x\begin{array}{|ccc|}2&3&1\\1&5&1\\3&2&1\end{array}-y\begin{array}{|ccc|}1&3&1\\2&5&1\\4&2&1\end{array}+z\begin{array}{|ccc|}1&2&1\\2&1&1\\4&3&1\end{array}-\begin{array}{|ccc|}1&2&3\\2&1&5\\4&3&2\end{array}=0$$
Compare this with $P:Ax+By+Cz=D$. So each of the above 3x3 determinant is a coefficient of the equation of the plane. 
Since you're given $M(1,2,3)$, $N(2,1,5)$ and  $P(4,3,2)$, you can easily find out that $\vec{MN}=(1,-1,2)$ and $\vec{MP}=(3,1,-1)$. The normal vector to the plane is then $\vec{n}=\vec{MN}\times \vec{MP}=(-1,7,4)$. Define an arbitrary point $K(x,y,z)$ that is on the plane.
Expand $\Delta$: $$\Delta=x[2(5-2)-3(1-3)+1(2-3\cdot 5)]-y[(5-2)-3(2-4)+(4-20)]+z[(1-3)-2(2-4)+(6-4)]-[(2-15)-2(4-20)+3(6-4)]=0\\\implies-x+7y+4z-25=0\\\iff -1(x-1)+7(y-2)+4(z-3)=0\\\iff \vec{n}\,\bullet\vec{MK}=0.$$
This means the normal vector $\vec{n}$ defined by the cross product of two vectors from 3 points is orthogonal to arbitrary vectors on the plane. Calculation of $\Delta$ addresses these reasoning.
