Let (P) be the plane with equation
$$ax+by+cz=0$$
(I have dropped the $d$ but it can easily been included).
$N(a,b,c)$ be a normal vector to (P); we will assume WLOG that $\|N\|^2=a^2+b^2+c^2=1$.
Suppose one wants to reflect a surface with equation
$$f(x,y,z)=0\tag{1}$$
for example the sphere with center $(1,0,0)$ and unit radius
$$(x-1)^2+y^2+z^2-1=0\tag{2}$$
The symmetry $X \leftrightarrow X'$ with respect to plane (P) is obtained through the equation (see explanation below):
$$X'=(I_3-2 NN^T)X \ \ \iff \ \ X=(I_3-2 NN^T)X' \tag{3}$$
(because symmetry is involutive ; N is assumed a column vector)
i.e., detailed in terms of coordinates:
$$\begin{pmatrix}x\\
y\\
z \end{pmatrix}=\begin{pmatrix}(1-2a^2)& -2ab&-2ac\\
-2ab&(1-2b^2)&-2bc \\
-2ac&-2bc&(1-2c^2)& \end{pmatrix}\begin{pmatrix}x'\\
y'\\
z'\end{pmatrix}\tag{4}$$
It is enough now to replace $x,y,z$ in (1) by their expressions issued from (4) to get the equations of the symmetric surface.
For example, if you reflect the sphere with equation (2) into vertical plane with equation $x=0$ (meaning that $a=1$, $b=c=0$), you will get $x=-x',y=y',z=z'$, giving:
$$(x'+1)^2+y'^2+z'^2=1$$
Remark: If you want to symmetrize a straight line, consider it classically as the intersection of two planes, and symmetrize separately the equations of two planes using the method just seen.
Explanation of formula (3):
the image of $X=N$ is $(I_3-2NN^T)N=N-2N\underbrace{(N^TN)}_{\|N\|^2=1}=-N$ as expected.
if $M \perp N$, $(I_3-2NN^T)M=M-2N\underbrace{(N^TM)}_0=M$.
Edit: (following a remark of @mr_e_man):
in fact, one can "mirror" any geometrical object with respect to a circle in 2D or a sphere in 3D using a transformation called "inversion" which is described here. The mirroring property can be seen on the example of the fish "reflected" with respect to his bowl. A circle is in general reflected into a circle. For extended information and references, see there.
