To find \( h \) for the function 

\(y = 1 + hx - 3x^2\)

with a minimum value of 13:

1. The vertex \( x \) is given by 

\(x = \frac{h}{6}\)

2. Substitute into the function:

\(y = 1 + \frac{h^2}{6} - 3\left(\frac{h}{6}\right)^2 = 1 + \frac{h^2}{6} - \frac{h^2}{12}\)

3. Combine terms:

\(y = 1 + \frac{2h^2}{12} - \frac{h^2}{12} = 1 + \frac{h^2}{12}\)

4. Set equal to 13:

\(1 + \frac{h^2}{12} = 13 \implies \frac{h^2}{12} = 12 \implies h^2 = 144 \implies h = 12 \text{ or } -12\)

\( h = 12 \) or \( h = -12 \).