Let's denote the amount of liquid in the first vat as \( V_1 \) and the amount in the second vat as \( V_2 \).
Initially, we have:
- \( V_1 = 121212 \) gallons (currently filling)
- \( V_2 = 100100100 \) gallons (currently leaking)
The first vat is being filled at a rate of \( 333 \) gallons per minute, meaning that after \( t \) minutes, the volume of liquid in the first vat will be:
\[
V_1(t) = 121212 + 333t
\]
The second vat is leaking at a rate of \( 777 \) gallons per minute, so after \( t \) minutes, the volume of liquid in the second vat will be:
\[
V_2(t) = 100100100 - 777t
\]
We want to find the time \( t \) when both vats have the same amount of liquid, i.e., when \( V_1(t) = V_2(t) \):
\[
121212 + 333t = 100100100 - 777t
\]
Now, we'll solve this equation for \( t \):
1. Rearranging the equation:
\[
333t + 777t = 100100100 - 121212
\]
2. Combine the terms:
\[
1110t = 100100100 - 121212
\]
Calculating the right-hand side:
\[
100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 99978888
\]
3. Now, substitute back into the equation:
\[
1110t = 99978888
\]
4. Solve for \( t \):
\[
t = \frac{99978888}{1110}
\]
5. Performing the division:
\[
t \approx 90099.036036
\]
This is in minutes. To convert that to seconds, we multiply by \( 60 \):
\[
t_{seconds} \approx 90099.036036 \times 60 \approx 5405942.1616
\]
So, rounding to the nearest whole number:
\[
t_{seconds} \approx 5405942
\]
Thus, both vats will have the same amount of liquid after approximately **5405942 seconds**.