There are two different vats. The first vat has 202020 gallons is being filled at a rate of 333 gallons per minute. The second vat has 116116116 gallons and is leaking fluid at a rate of 222 gallons per minute. After how many seconds will both vats have the same amount of liquid

Let \( V_1(t) \) be the volume of liquid in the first vat at time \( t \) (in minutes), where it starts with 0 gallons and is filling at a rate of 333 gallons per minute. The volume of the first vat at time \( t \) is given by:

\[
V_1(t) = 333t
\]

The first vat has a capacity of 202020 gallons. Therefore, the volume reaches its maximum when:

\[
333t = 202020 \implies t = \frac{202020}{333} \approx 606.06 \text{ minutes}
\]

At that point, the first vat will be at its full capacity.

Now, let \( V_2(t) \) be the volume of liquid in the second vat at time \( t \). The second vat starts with 116116116 gallons and is leaking at a rate of 222 gallons per minute, so we have:

\[
V_2(t) = 116116116 - 222t
\]

Next, we need to find the time \( t \) when both vats have the same volume of liquid:

\[
V_1(t) = V_2(t)
\]

Substituting the equations we have:

\[
333t = 116116116 - 222t
\]

Combining terms:

\[
333t + 222t = 116116116
\]

\[
555t = 116116116
\]

Solving for \( t \):

\[
t = \frac{116116116}{555} \approx 209,839.29 \text{ minutes}
\]

Now, to convert this to seconds:

\[
t = 209839.29 \times 60 \approx 12591057.4 \text{ seconds}
\]

Thus, the time at which both vats will have the same amount of liquid is approximately \( \boxed{12591057.4} \) seconds.