Find the coordinate of third point of equilateral triangle.

Call the position of point $C$ by the coords $(a, b)$. The equations for $C$ are then

$$ \sqrt{(a-3)^2 + (b - 4)^2} = \sqrt{26} \\ \sqrt{(a+2)^2 + (b - 3)^2} = \sqrt{26} $$ Squaring both, we get $$ (a-3)^2 + (b - 4)^2 = 26 \\ (a+2)^2 + (b - 3)^2 = 26 $$ $$ a^2 - 6a + 9 + b^2 - 8b + 16= 26 \\ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ Subtracting these two gives $$ -10a + 5 - 2b + 7 = 0 $$ or $$ 6 = 5a + b $$ which is a line both points must lie on. Writing this as $$ b = 6 - 5a $$ we can substitute in either equation. Let's got with the second:

$$ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ becomes $$ a^2 + 4a + 4 + (6-5a)^2 - 6(6-5a) + 9= 26 $$ which is a quadratic that can now be solved for the two possible values of $a$.

(Once you do so, you use $b = 6 - 5a$ to find the corresponding $b$-values.)

If you like vector approach:

Displace (shift/translate) point A by vector(-3,-4) to come to the origin.

Multiply the new radius vector $ab$ with $ e^ {i \pi/3} , e^ {-i \pi/3}$ (once clockwise and once anticlockwise) to obtain new points $ C_1 $ and $ C_2 $. The multiplying factor is $ (1/2 \pm i \sqrt 3/2) $.

Displace these points back to original positions translating by $ (3,4) $.

If you multiply thrice, all points of a hexagon would also be reached.