Step-by-Step Solution
Key Concept: A family of lines passes through a fixed point when the coefficients of parameters can be independently set to zero.
The given line is $aα + by + c - 1 = 0$. Simplifying: $20ax + 20by + 20c - 20 = 0$. From the relation $20ax + 20by + t - 5a - 4b - 20 = 0$, we substitute and rearrange to get $20a(x - \frac{1}{4}) + 20b(y - \frac{1}{5}) + (t - 20) = 0$. For this to hold for all values of $a$ and $b$, the line must pass through the point $(\frac{1}{4}, \frac{1}{5})$ for $t = 20$. Therefore, the value is $t = 20$.
Correct Answer: 20