$F_1 = 6(220) = 1320 \, \text{ N}$
$F_2 = \frac{1}{2}(6)(670) = 2010 \, \text{ N}$
$R = F_1 + F_2 = 1320 + 2010$
$R = 3330 \, \text{ N}$
$Rd = 3F_1 + 4F_2$
$3330d = 3(1320) + 4(2010)$
$3330d = 12000$
$d = 3.6 \, \text{ m}$
Thus, R = 3330 N downward at 3.6 m to the right of A. answer