Translate ABC using the vector -3 1 to obtain A'B'C'.

Here are the steps for the composite transformation:

Step 1: Translate $\triangle ABC$ using the vector $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$ to obtain $\triangle A'B'C'$.
The coordinate rule for this translation is $(x, y) \to (x-3, y+1)$.
Applying the rule to the vertices $A(1, 2)$, $B(3, 4)$, and $C(2, 0)$:

$$A' = (1-3, 2+1) = (-2, 3)$$ $$B' = (3-3, 4+1) = (0, 5)$$ $$C' = (2-3, 0+1) = (-1, 1)$$

The coordinates of $\triangle A'B'C'$ are $A'(-2, 3)$, $B'(0, 5)$, and $C'(-1, 1)$.

Step 2: Reflect $\triangle A'B'C'$ in the y-axis to obtain $\triangle A''B''C''$.
The coordinate rule for reflection in the y-axis is $(x, y) \to (-x, y)$.
Applying the rule to the vertices $A'(-2, 3)$, $B'(0, 5)$, and $C'(-1, 1)$:

$$A'' = (-(-2), 3) = (2, 3)$$ $$B'' = (-(0), 5) = (0, 5)$$ $$C'' = (-(-1), 1) = (1, 1)$$

The coordinates of $\triangle A''B''C''$ are $A''(2, 3)$, $B''(0, 5)$, and $C''(1, 1)$.

Step 3: Rotate $\triangle A''B''C''$ through $90^\circ$ anticlockwise about the origin to obtain $\triangle A'''B'''C'''$.
The coordinate rule for a $90^\circ$ anticlockwise rotation about the origin is $(x, y) \to (-y, x)$.
Applying the rule to the vertices $A''(2, 3)$, $B''(0, 5)$, and $C''(1, 1)$:

$$A''' = (-(3), 2) = (-3, 2)$$ $$B''' = (-(5), 0) = (-5, 0)$$ $$C''' = (-(1), 1) = (-1, 1)$$

The coordinates of $\triangle A'''B'''C'''$ are $A'''(-3, 2)$, $B'''(-5, 0)$, and $C'''(-1, 1)$.

Step 4: Enlarge $\triangle A'''B'''C'''$ with a scale factor of $k=2$ centred at the origin to obtain $\triangle A''''B''''C''''$.
The coordinate rule for enlargement with scale factor $k=2$ centred at the origin is $(x, y) \to (2x, 2y)$.
Applying the rule to the vertices $A'''(-3, 2)$, $B'''(-5, 0)$, and $C'''(-1, 1)$:

$$A'''' = (2 \times -3, 2 \times 2) = (-6, 4)$$ $$B'''' = (2 \times -5, 2 \times 0) = (-10, 0)$$ $$C'''' = (2 \times -1, 2 \times 1) = (-2, 2)$$

The coordinates of $\triangle A''''B''''C''''$ are $A''''(-6, 4)$, $B''''(-10, 0)$, and $C''''(-2, 2)$.

To create the labeled graph, plot the original triangle $ABC$ and all the transformed images $A'B'C'$, $A''B''C''$, $A'''B'''C'''$, and $A''''B''''C''''$ using the coordinates calculated above on a coordinate plane.

Step 5: Compare the perimeter of $\triangle ABC$ and $\triangle A''''B''''C''''$.
First, calculate the side lengths of $\triangle ABC$ using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:

$$AB = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8}$$ $$BC = \sqrt{(2-3)^2 + (0-4)^2} = \sqrt{(-1)^2 + (-4)^2} = \sqrt{1+16} = \sqrt{17}$$ $$CA = \sqrt{(1-2)^2 + (2-0)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$$

The perimeter of $\triangle ABC$ is $P_{ABC} = AB + BC + CA = \sqrt{8} + \sqrt{17} + \sqrt{5}$.

The final transformation from $\triangle A'''B'''C'''$ to $\triangle A''''B''''C''''$ is an enlargement with a scale factor of $k=2$. All previous transformations (translation, reflection, rotation) are rigid transformations, meaning they preserve length and shape. Therefore, $\triangle A'''B'''C'''$ has the same perimeter as $\triangle ABC$.
When a shape is enlarged by a scale factor $k$, its perimeter is multiplied by $k$.
So, the perimeter of $\triangle A''''B''''C''''$ is $k$ times the perimeter of $\triangle ABC$.

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