The given inequality is $x + \large\frac{x}{2} +\frac{x}{3} $$< 11$
$=> x [1+ \large\frac{1}{2} +\frac{1}{3} \bigg] $$<11$
$=> x \bigg[ \large\frac{6+ 3+2}{6} \bigg]$$ <11$
$=> \large \frac{11 x}{6} $$<11$
Dividing both sides of the inequality by 11.
=> $ \large\frac{11 x}{6 \times 11} < \frac{11}{11}$
=> $ \large\frac{x}{6} $$<1$
=> $ x < 6$
Step 2:
All real numbers which are less than 6 satisfy the given in equality.
The solution set is $(-\infty, 6)$
Hence B is the correct answer.