Three points A , B , and C lie on a straight line on a horizontal...

Let the height of the vertical pole $OP$ be $h$.

The horizontal distances from the base of the pole $O$ to the points $A$, $B$, and $C$ can be found using the angles of elevation:
$OA = h \cot 30^\circ = h\sqrt{3}$
$OB = h \cot 45^\circ = h$
$OC = h \cot 60^\circ = \dfrac{h}{\sqrt{3}}$

In $\triangle OAC$, since $B$ is the midpoint of $AC$, $OB$ is the median to the side $AC$. We can apply Apollonius' Theorem:
$OA^2 + OC^2 = 2\left(OB^2 + \left(\dfrac{AC}{2}\right)^2\right)$

Substitute the values into the theorem:
$(h\sqrt{3})^2 + \left(\dfrac{h}{\sqrt{3}}\right)^2 = 2\left(h^2 + 10^2\right)$

$3h^2 + \dfrac{h^2}{3} = 2(h^2 + 100)$

$\dfrac{10h^2}{3} = 2h^2 + 200$

$\dfrac{10h^2}{3} - 2h^2 = 200$

$\dfrac{4h^2}{3} = 200$

$h^2 = \dfrac{200 \times 3}{4} = 150$

The square of the height of the pole is $150$ m$^2$.

Answer: $150$