The resultant of three vectors 1, 2, and 3 units whose directions are those of the sides of an equilateral triangle is at an angle of

Analytically:

${\mathrm{R}}_{\mathrm{x}}$ = 1 + 2 cos 120° + 3 cos 240°

= -3/2 

${\mathrm{R}}_{\mathrm{y}}$ = 2 sin 120° + 3 sin 240° 

$=2\times \frac{\sqrt{3}}{2}+3\times \left(-\frac{\sqrt{3}}{2}\right)=-\frac{\sqrt{3}}{2}$

$\mathrm{tan\theta }=\frac{{\mathrm{R}}_{\mathrm{y}}}{{\mathrm{R}}_{\mathrm{x}}}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{3}{2}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\Rightarrow \mathrm{\theta }=30°$

Hence, angle with first vector is 180 - $\mathrm{\theta }$ = 150°. 
Graphically:

AB = 1, BC = 2, CD = 3
$\overrightarrow{\mathrm{R}}=\overrightarrow{\mathrm{AD}}$ is resultant, clearly angle between  $\overrightarrow{\mathrm{R}}$ and $\overrightarrow{\mathrm{AB}}$ is 150°.