The problem states that the figure shows the graph of the quadratic equation:

$y = 3x^2 - 6x + (k - 2)$

with the axis of symmetry at $x = 1$. Let's analyze the situation:

Step 1: General Formula for Axis of Symmetry

For a quadratic equation of the form: $y = ax^2 + bx + c,$ the axis of symmetry is given by: $x = -\frac{b}{2a}.$

In this case: $a = 3, \quad b = -6, \quad c = k - 2.$

Step 2: Plugging in the Axis of Symmetry

The problem specifies the axis of symmetry is at $x = 1$. Thus: $1 = -\frac{-6}{2(3)}.$

This confirms that the quadratic is correctly set up, as the axis of symmetry depends only on $a$ and $b$, and is independent of $c$.

Step 3: Determine the Vertex

The axis of symmetry passes through the vertex of the parabola. Substituting $x = 1$ into the quadratic equation: $y = 3(1)^2 - 6(1) + (k - 2).$ Simplifying: $y = 3 - 6 + (k - 2),$ $y = k - 5.$

Thus, the vertex is at: $(1, k - 5).$

Step 4: Analyze the Graph and Solve for $k$

From the graph, it appears that the vertex lies on the horizontal axis ($y = 0$). Hence: $k - 5 = 0,$ $k = 5.$

Final Answer:

The value of $k$ is: $\boxed{5}.$

Would you like further clarification or details on this? Here are five related questions to deepen understanding:

1. How do we determine the axis of symmetry for any quadratic equation?
2. Why does the vertex always lie on the axis of symmetry?
3. What is the significance of the $k - 2$ term in this quadratic equation?
4. How do the graph’s roots relate to its vertex and axis of symmetry?
5. How does changing $k$ affect the graph of the quadratic function?

Tip: Always double-check the graph's key points, like the vertex and axis of symmetry, to confirm your solutions!