Explanation

(d) The three vectors $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ are represented as shown in figure (a) where $A = 1,B = 2$ and $C = 3$. Here the sides of the equilateral triangle represent only the directions and not the magnitudes of the vectors.
In figure (b), these vector are drawn from a common point O and they are lying in $X - Y$ plane. Resolving these vectors into two rectangular components along $X$-axis and $Y$-axis, we have, the $X$-component of resultant vector as
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$R_{X} = \left| \overrightarrow{A} \right| + \left| \overrightarrow{B} \right|$ cos $(180{^\circ} - 60{^\circ}) + \left| \overrightarrow{C} \right|\cos(180{^\circ} + 60{^\circ})$
$= - 1 - 2cos60{^\circ} - 3cos60{^\circ}$
$= - 1 - 2 \times \frac{1}{2} - 3 \times \frac{1}{2} = - \frac{3}{2}$
$Y$-component of resultant vector is
$R_{Y} = 0 + |B|\sin(180{^\circ} - 60{^\circ}) + \left| \overrightarrow{C} \right|\sin(180{^\circ} + 60{^\circ})$
$= 0 + 2sin60{^\circ} - 3sin60{^\circ} = - sin60{^\circ} = \sqrt{3}/2$
Magnitude of resultant vector,
$R = \sqrt{R_{X}^{2} + R_{Y}^{2}} = \sqrt{\left( - \frac{3}{2} \right)^{2} + \left( - \frac{\sqrt{3}}{2} \right)^{2}}$
$= \sqrt{3}$ units