3. The graph of the equation $6 x^{2}-5 x y+y^{2}-8 x+2 y-8=0$ and the $x$-axis enclose a figure whose area is $\qquad$
3. $\frac{64}{3}$.

Given $(2 x-y-4)(3 x-y+2)=0$.
Therefore, the image of the equation with the $x$-axis forms a triangle.
$$
\text { Let }\left\{\begin{array}{l}
2 x-y-4=0, \\
3 x-y+2=0,
\end{array}\right.
$$

The intersection point coordinates are $(-6,-16)$.
The $x$-intercepts of the lines $2 x-y-4=0$ and $3 x-y+2=0$ are $2$ and $-\frac{2}{3}$, respectively. Thus, the area of the triangle is
$$
\frac{1}{2}\left(2+\frac{2}{3}\right) \times 16=\frac{64}{3} \text {. }
$$