The base of a triangular prism is $ABC$. $A'B'C'$ is an equilateral triangle with lengths $a$... The base of a triangular prism is $ABC$. $A'B'C'$ is an equilateral triangle with lengths $a$, and the lengths of its adjacent sides also equal $a$. Let $I$ be the midpoint of $AB$ and $B'I \perp (ABC)$. Find the distance from the $B'$ to the plane ($ACC'A'$) in term of $a$.

Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $\angle{ABC}$ will be $\pi/3$ (as $\Delta BIB'$ is a $(1,\sqrt{3},2)$ right-triangle). The co-ordinates of $A$ will then be of the form $\left(a\cos{\pi/3},a\cos{\pi/3},h\right)$. As the length of $AB$ is $a$, it leads to 
$$
\left(a\cos{\pi/3}\right)^{2}+\left(a\cos{\pi/3}\right)^{2}+h^{2}=a^{2} \\
h=\frac{a}{\sqrt{2}}
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/\sqrt{2}$.
