U are forgetting that z equation

You are right to point out the curve equation, Desmond. Let's clarify the situation.

The problem states that a point is moving along the intersection of two equations:

1. Curve: $x^2 + 3xy + 3y^2 = 22$
2. Line: $x - 2y + 4 = 0$

We are asked to find the speed of this point when $x=2$, given $\frac{dx}{dt} = 3$ units/second.

First, let's check if there is a point on the intersection where $x=2$.
If $x=2$, from the line equation $x - 2y + 4 = 0$:

$$2 - 2y + 4 = 0$$ $$6 - 2y = 0$$ $$2y = 6$$ $$y = 3$$

So, the point $(2, 3)$ lies on the line.

Now, let's check if this point $(2, 3)$ also lies on the curve $x^2 + 3xy + 3y^2 = 22$:

$$(2)^2 + 3(2)(3) + 3(3)^2 = 4 + 18 + 27 = 49$$

Since $49 \neq 22$, the point $(2, 3)$ is not on the curve.

This means there is no point that satisfies both the curve equation and the line equation at $x=2$. Therefore, a point cannot be "moving along the intersection" when $x=2$. The problem statement contains an inconsistency.

To provide a solution, we must make a reasonable assumption. The most common approach in such cases is to assume the problem intends for the point to be moving along one of the given paths, and the other information is either extraneous or there's a typo in the problem. Given that the line equation directly relates $x$ and $y$ in a simpler way, we will assume the point is moving along the line $x - 2y + 4 = 0$ at the instant $x=2$.