The equation of the axis of symmetry for a parabola given in the vertex form \( y = a(x - h)^2 + k \) is \( x = h \), where \( (h, k) \) is the vertex of the parabola.
In your equation \( y = 3(x - 1)^2 - 12 \), the vertex can be identified as \( (h, k) \) where \( h = 1 \) and \( k = -12 \).
Therefore, the equation for the axis of symmetry is:
\[ x = 1 \]