6. In the Cartesian coordinate system $x O y$, it is known that points $A(6,4)、B(4,0)、C(0,3)$, and the line $l: y=k x$ bisects the area of $\triangle A B C$, $k=\frac{-b+\sqrt{c}}{a}(a、b、c$ are all positive integers, and $c$ has no square factors). Then $a+b+c=$ $\qquad$ .
6.584.

As shown in Figure 5, take the midpoint $M\left(2, \frac{3}{2}\right)$ of side $BC$. Let the line $l$ intersect sides $BC$ and $CA$ at points $P$ and $Q$, respectively.

Since $AM$ and $PQ$ both bisect the area of $\triangle ABC$, we have $S_{\triangle PQC} = S_{\triangle MMC}$. Subtracting the common area $S_{\triangle CMQ}$, we get $S_{\triangle PMQ} = S_{\triangle \triangle MQ}$. Therefore, the distances from points $P$ and $A$ to the line $MQ$ are equal. Hence, $PA \parallel MQ$.
The equation of line $l_{BC}$ is $\frac{x}{4} + \frac{y}{3} = 1$, or $y = 3 - \frac{3}{4}x$;
The equation of line $l_{AC}$ is $\frac{y-3}{x-0} = \frac{4-3}{6-0}$, or $y = \frac{1}{6}x + 3$;
The equation of line $l$ is $y = kx$.
From equations (1) and (3), we get $P\left(\frac{12}{4k+3}, \frac{12k}{4k+3}\right)$,
From equations (2) and (3), we get $Q\left(\frac{18}{6k-1}, \frac{18k}{6k}-1\right)$.
Since $PA \parallel MQ$, we have $k_{PA} = k_{MQ}$, leading to $132k^2 + 77k - \frac{231}{2} = 0$.
Thus, $k = \frac{-7 + \sqrt{553}}{24}$.
Therefore, $a + b + c = 24 + 7 + 553 = 584$.