Let $t$ be a parameter that varies over all real numbers.  Any parabola of the form
\[y = 3x^2 + tx - 2t\]passes through a fixed point.  Find this fixed point.
To obtain the fixed point, we want to eliminate $t$ in the equation
\[y = 3x^2 + tx - 2t.\]We can do so by taking $x = 2.$  This leaves us with $y = 3 \cdot 2^2 = 12,$ so the fixed point is $\boxed{(2,12)}.$
(2,12)
Level 3
Intermediate Algebra
Problem: Let $t$ be a parameter that varies over all real numbers.  Any parabola of the form
\[y = 3x^2 + tx - 2t\]passes through a fixed point.  Find this fixed point.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the fixed point through which all parabolas of the form y = 3x^2 + tx - 2t pass, we need to analyze the equation. The parameter t affects the linear term and the constant term of the parabola. 

We can express the condition for a point (a, b) to lie on the parabola for all values of t. Substituting (a, b) into the equation gives us:

b = 3a^2 + ta - 2t.

Rearranging this equation, we can isolate the terms involving t:

b = 3a^2 + t(a - 2).

For this equation to hold for all values of t, the coefficient of t must equal zero. This leads to the condition:

a - 2 = 0.

Solving this gives us a = 2. 

Next, we substitute a = 2 back into the equation to find b:

b = 3(2^2) + t(2) - 2t.

This simplifies to:

b = 3(4) + 2t - 2t,
b = 12.

Thus, the fixed point that all parabolas pass through is (2, 12).<|END_OF_LATENT|> The final answer is $(2,12)$.