55. Points $A$ and $A^{\prime}$ are given by their coordinates: $(7,1,4)$ and $(3,5,2)$. Find the equation of the plane relative to which these points are symmetric.
55. All points with coordinates ( $x, y, z$ ), equally distant from two given points, satisfy the conditions:

$$
(x-7)^{2}+(y-1)^{2}+(z-4)^{2}=(x-3)^{2}+(y-5)^{2}+\cdot(z-2)^{2} .
$$

Expanding the brackets, we get:

$$
\begin{gathered}
14 x-49+2 y-1+8 z-16=6 x-9+10 y-25+4 z-4 \\
8 x-8 y+4 z-28=0 ; 2 x-2 y+z-7=0
\end{gathered}
$$

We will bring the equation of the plane to its normal form. Since $\sqrt{2^{2}+2^{2}+1}=3$, the normal equation will take the form: $\frac{2}{3} x$ $-\frac{2}{3} y+\frac{z}{3}-\frac{7}{3}=0$