Point $A$ is a point on the line $l: y=3 x$, and it is in the first quadrant. Point $B$ has coordinates $(3,2)$, and the line $A B$ intersects the $x$-axis at point $C$. What is the minimum value of the area of triangle $A O C$?
(A) $\frac{27}{2}$
(B) $\frac{28}{3}$
(C) $\frac{793}{66}$
(D) $\frac{93}{19}$
6. (B).

If $A C: x=3$, then $A(3,9)$, at this time $S_{\text {oct.ux }}=\frac{27}{2}$.
Let $A C: y-2=k(x-3)$, then $C\left(-\frac{2}{k}+3,0\right)$, solving the system of equations $\left\{\begin{array}{l}y-2=k(x-3), \\ y=3 x\end{array}\right.$ we get $A\left(\frac{3 k-2}{k-3}, 3 \cdot \frac{3 k-2}{k-3}\right)$, since $\mathrm{A}$ is in the first quadrant, thus $\frac{3 k-2}{k-3}>0$, so $\frac{2}{3}<k<3$, let $u=\frac{9}{k}$, then $k=\frac{9}{u}$, substituting into the equation of line $A C$ we get $(9-u) k^{2}+(3 u-12) k+4=0, \Delta=9 u^{2}-56 u \geqslant 0$, since $u \geqslant 0$, so $u \geqslant \frac{56}{9}$, thus $S_{\triangle A O C} \geqslant \frac{3}{2} \times \frac{56}{9}=\frac{28}{3}\left(<\frac{27}{2}\right)$.
Therefore, the minimum value of the area of $\triangle A O C$ is $\frac{28}{3}$.