Three charges are placed at the vertices of an equilateral triangle...

The system of charges can be considered as two electric dipoles by splitting the $2q$ charge into two $+q$ charges.

Let the vertices be $A$ (top, $-q$), $B$ (bottom-left, $2q$), and $C$ (bottom-right, $-q$).

The first dipole is formed by $-q$ at $A$ and $+q$ at $B$. Its dipole moment $\vec{p}_1$ has magnitude $qd$ and is directed from $A$ to $B$.

The second dipole is formed by $-q$ at $C$ and $+q$ at $B$. Its dipole moment $\vec{p}_2$ has magnitude $qd$ and is directed from $C$ to $B$.

The angle between the vectors $\vec{p}_1$ and $\vec{p}_2$ is $60^\circ$, which is the interior angle of the equilateral triangle at vertex $B$.

The resultant dipole moment $\vec{P}$ is the vector sum of $\vec{p}_1$ and $\vec{p}_2$.

The magnitude of the resultant dipole moment is given by:
$P = \sqrt{p_1^2 + p_2^2 + 2p_1 p_2 \cos(60^\circ)}$
$P = \sqrt{(qd)^2 + (qd)^2 + 2(qd)^2 \left(\dfrac{1}{2}\right)}$
$P = \sqrt{3(qd)^2} = \sqrt{3}qd$

Since the magnitudes of $\vec{p}_1$ and $\vec{p}_2$ are equal, the resultant vector $\vec{P}$ lies along the bisector of the angle between them.

Both $\vec{p}_1$ and $\vec{p}_2$ point towards vertex $B$. Therefore, their vector sum points from the inside of the triangle towards $B$ and continues outwards. This direction is along the bisector of the angle at $2q$, pointing away from the triangle.

Answer: $\sqrt{3} qd$, along the bisector of the angle at $2q$, away from the triangle