side length.

Place the vertices as:

$$ A(a,0),\quad B\left(\frac a2,\frac{\sqrt3a}{2}\right),\quad C\left(-\frac a2,\frac{\sqrt3a}{2}\right) $$

$$ D(-a,0),\quad E\left(-\frac a2,-\frac{\sqrt3a}{2}\right),\quad F\left(\frac a2,-\frac{\sqrt3a}{2}\right) $$

Now,

$$ \overrightarrow{AB}=B-A=\left(-\frac a2,\frac{\sqrt3a}{2}\right) $$

$$ \overrightarrow{AC}=C-A=\left(-\frac{3a}{2},\frac{\sqrt3a}{2}\right) $$

$$ \overrightarrow{AD}=D-A=(-2a,0) $$

$$ \overrightarrow{AE}=E-A=\left(-\frac{3a}{2},-\frac{\sqrt3a}{2}\right) $$

$$ \overrightarrow{AF}=F-A=\left(-\frac a2,-\frac{\sqrt3a}{2}\right) $$

Adding them,

$$ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF} $$

The $y$-components cancel each other, and the $x$-components give:

$$ -\frac a2-\frac{3a}{2}-2a-\frac{3a}{2}-\frac a2=-6a $$

So,

$$ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=(-6a,0) $$

Now,

$$ \overrightarrow{AD}=(-2a,0) $$

Hence,

$$ (-6a,0)=3\overrightarrow{AD} $$

So,

$$ p=3 $$

Also,

$$ \overrightarrow{AO}=O-A=(-a,0) $$

So,

$$ (-6a,0)=6\overrightarrow{AO} $$

Hence,

$$ q=6 $$

So the values of $p$ and $q$ are:

$$ \boxed{3,6} $$

Correct option: C.

Vector Algebra

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