I'm asked to determine the Cartesian equation for a plane through a point $(2, 3, 1)$ and that is at equal distance to three different points $A(1, 3, 5)$, $B(1, 1, 1)$, $C(3, 1, 3)$. I have the formula for the distance between a point and a plane:
$$ d(\vec{p}, V) = \frac{\mid d - ax_0 - by_0 - cz_0\mid}{\sqrt{a^2 + b^2 + c^2}}$$.
I have also introduced a new variable, $\lambda$, to represent the distance between the plane and the point. Thus I get the system of equations:
$$\begin{cases} \lambda = d(\vec{p_a}, V) = \frac{\mid d -a - 3b - 5c\mid}{\sqrt{a^2 + b^2 + c^2}}\\ \lambda = d(\vec{p_b}, V) = \frac{\mid d -a - b - c\mid}{\sqrt{a^2 + b^2 + c^2}}\\ \lambda = d(\vec{p_c}, V) = \frac{\mid d -3a - b - 3c\mid}{\sqrt{a^2 + b^2 + c^2}}\\ \end{cases}$$
I have no idea if this is the right start or not, but this is what I went with. But now I'm stuck and I have no idea on how to continue.
Let the equation of the plane be $(x-2)+p(y-3)+q(z-1)=0$ that contains the point $(2,3,1)$. Then, use the point-plane distance formula to establish
$$\frac{|-1+4q|}d = \frac{|-1-2p|}d = \frac{|-1-2p+2q|}d $$
with $d=\sqrt{1+p^2+q^2}$. Solve for $p$ and $q$ to obtain the planes below
\begin{align} & x=2\\ & x+ 2y -z= 7\\ & x -3y-z= -8\\ & 3x+y+2z=11 \end{align}