Plot the triangle ABC.

The coordinates of triangle ABC are given as:
A$(1,2)$
B$(1,6)$
C$(5,5)$

The rule for reflection across the line $y=x$ is $(x,y) \to (y,x)$.
Applying this rule to the vertices of triangle ABC:
A$(1,2) \to$ A'$(2,1)$
B$(1,6) \to$ B'$(6,1)$
C$(5,5) \to$ C'$(5,5)$
The coordinates of the reflected triangle A'B'C' are:
A'$(2,1)$
B'$(6,1)$
C'$(5,5)$

The rule for reflection across the y-axis is $(x,y) \to (-x,y)$.
Applying this rule to the vertices of triangle ABC:
A$(1,2) \to$ A''$(-1,2)$
B$(1,6) \to$ B''$(-1,6)$
C$(5,5) \to$ C''$(-5,5)$
The coordinates of the reflected triangle A''B''C'' are:
A''$(-1,2)$
B''$(-1,6)$
C''$(-5,5)$

The rule for rotation by $-180^\circ$ (or $180^\circ$) about the origin is $(x,y) \to (-x,-y)$.
Applying this rule to the vertices of triangle A'B'C' (from Step 2):
A'$(2,1) \to$ A'''$(-2,-1)$
B'$(6,1) \to$ B'''$(-6,-1)$
C'$(5,5) \to$ C'''$(-5,-5)$
The coordinates of the rotated triangle A'''B'''C''' are:
A'''$(-2,-1)$
B'''$(-6,-1)$
C'''$(-5,-5)$

First, let's find the combined transformation from ABC to A'''B'''C'''.
The transformation from ABC to A'B'C' is reflection across $y=x$: $(x,y) \to (y,x)$.
The transformation from A'B'C' to A'''B'''C''' is rotation by $-180^\circ$ about the origin: $(x,y) \to (-x,-y)$.
Combining these, for an original point $(x,y)$ from ABC:

1. Reflect across $y=x$: $(x,y) \to (y,x)$
2. Rotate by $-180^\circ$ about the origin: $(y,x) \to (-y,-x)$ So, the overall transformation from ABC to A'''B'''C''' is $(x,y) \to (-y,-x)$.

To determine if this is a reflection and find the mirror line, we check two conditions:

1. The midpoint of a point and its image must lie on the mirror line.
2. The line segment connecting a point and its image must be perpendicular to the mirror line.

Let a point be $(x,y)$ and its image be $(-y,-x)$.
The midpoint M is $\left(\frac{x+(-y)}{2}, \frac{y+(-x)}{2}\right) = \left(\frac{x-y}{2}, \frac{y-x}{2}\right)$.
Consider the line $y=-x$. If M lies on this line, then its y-coordinate must be the negative of its x-coordinate:
$\frac{y-x}{2} = -\left(\frac{x-y}{2}\right)$
$\frac{y-x}{2} = \frac{y-x}{2}$
This is true for all points, so the midpoint lies on the line $y=-x$.

The slope of the line connecting $(x,y)$ and $(-y,-x)$ is $m_1 = \frac{-x-y}{-y-x} = \frac