6. If the function of $x$
$$
y=(a-3) x^{2}-(4 a-1) x+4 a
$$

intersects the coordinate axes at two points, then the value of $a$ is
$\qquad$ .
ii. $6.3,0$ or $-\frac{1}{40}$.
When $a-3=0$, the original function becomes $y=-11 x+12$, whose graph intersects the coordinate axes at two points.

When $a-3 \neq 0$, the original function is a quadratic function, and its graph must intersect the $y$-axis at one point $(0,4a)$.

If this intersection point is not the origin, according to the given information, its graph can only intersect the $x$-axis at one point, so,
$$
\Delta=(4 a-1)^{2}-16 a(a-3)=0 .
$$

Solving this, we get $a=-\frac{1}{40}$.
If this intersection point is the origin, then $a=0$, in which case the function becomes $y=-3 x^{2}+x$, and its graph must intersect the $x$-axis at two different points.
In summary, the value of $a$ is 3, 0, or $-\frac{1}{40}$.