Regular tetrahedron Calculate the distance between the slanted wall heights of the regular tetrahedron with edge lengths $a$.
I have problem with visualization where the distance is. 
Let's say we have such situation($MC$ is height) and draw yet height $DP(BC)$, so the quested length is between $P$ and projected point $P$ on $MC$ ?

 A: I am trying to answer your question (with my own interpretation). Several lines have been added to your picture. See below.
First of all, because MC is on the base (according to your picture), it should not be used as 'slant height'. To certain extend, you are allowed to do so because the figure is regular.
I found two of the slant heights, one is DM on face DAB and the other is DP on face DBC. The 'P' that you are using is not very clear because it was not shown in your picture.
To find the required distance is impossible because there are too many answers (ranging from 0 to MP). I guess what you are looking for is the MAXIMUM distance between these two slant heights. If this is true, then the answer is MP, which is eventually equal to 0.5a  because triangle BMP (was mistakenly written as DMP) is equilateral.   
A: Let $M$ be the middle point of $AB$ (as in the picture), and $P$ be the middle point of $BC$. We shall find the distance between segments $CM$ and $DP$.
First, it is clear that the answer is proportional to $a$, so I will just assume that $a=1$ to make formulas shorter.
The plan is very simple. I will calculate lots of vectors, and for my calculations I will use vectors $\overline{BA}, \overline{BC}, \overline{BD}$ as a basis. In the calculations to come I will use dot products of these vectors, which I denote by $(,)$. Their lengths are all equal to $1$, so their products by themselves are $1$:
$$
(\overline{BA}, \overline{BA}) = 
(\overline{BC}, \overline{BC}) = 
(\overline{BD}, \overline{BD}) = 1.
$$
The pairwise dot products are equal to $1/2$, because the angle between each two of these vectors is 60 degrees, and its cosine is $1/2$. So we have
$$
(\overline{BA}, \overline{BC}) =
(\overline{BC}, \overline{BD}) = 
(\overline{BD}, \overline{BA}) = 1/2.
$$
Now, let $Q$ be a point on line $MC$ and $R$ be a point on line $PD$ such that the distance between $Q$ and $R$ is the smallest possible. We will see later that in fact $Q$ is on segment $MC$ and $R$ is on segment $PD$, so the length of $QR$ is the answer to our problem.
To find where $Q$ and $R$ are situated, let's use the parametric line equations in vector form:
$$
\begin{align}
\overline{BQ} &= \overline{BC} + \alpha\left(\overline{BC} - \frac{1}{2} \overline{BA}\right) \\
\overline{BR} &= \overline{BD} + \beta\left(\overline{BD} - \frac{1}{2} \overline{BC}\right)
\end{align}
$$
Here $\alpha$ and $\beta$ are some constants that we have to find. First, let's subtract these equations to get vector $\overline{QR}$:
$$
\overline{QR} = \overline{BR} - \overline{BQ} =
\frac{1}{2}\alpha\overline{BA} + \left(-1 - \alpha - \frac{1}{2}\beta\right)\overline{BC} + (1 + \beta)\overline{BD}.
$$
It should be clear that vector $\overline{QR}$ must be orthogonal to both lines $MC$ and $PD$. In terms of dot products this can be written as a system of two equations:
$$
\begin{align}
\left(\overline{QR}, \overline{BC} - \frac{1}{2}\overline{BA}\right) &= 0; \\
\left(\overline{QR}, \overline{BD} - \frac{1}{2}\overline{BC}\right) &= 0.
\end{align}
$$
Now, using our formula for $QR$ and the known dot products of our basis vectors, we can boil these equations down to the following:
$$
\begin{align}
6\alpha + \beta &= -4; \\
\alpha + 6\beta &= -6.
\end{align}
$$
Solving the system, we get $\alpha = -\frac{18}{35}$ and $\beta = -\frac{32}{35}$. The fact that both numbers are between $-1$ and $0$ shows that both points $Q$ and $R$ lie on their respective heights, not on their continuations.
And now the deed is almost done. We know $\alpha$ and $\beta$, so we can find
$$
\overline{QR} = -\frac{9}{35}\overline{BA} -\frac{1}{35}\overline{BC} + \frac{3}{35}\overline{BD}.
$$
To find its length, we find its dot product with itself (using the known dot products of the basis vectors). The result of this calculation is that $(\overline{QR}, \overline{QR}) = \frac{70}{35^2}$, therefore $|QR| = \frac{\sqrt{70}}{35}$. And that is our answer.
If we come back to an arbitrary value of $a$ instead of $1$, it is clear that the answer will be $\frac{\sqrt{70}}{35}a$.
I am really, really not in the mood to do the same amount of calculations for pair $(CM, BQ)$. Also, there surely is a shorter way to do this. Maybe calculations will be easier if you choose your basis better. I picked one at random and stuck with it. There should also be a solution without linear algebra, but somehow I don't feel like looking for it now. Maybe someone will provide one in another answer.
UPDATE: note that this method may look like an overkill, but there is an upside. It gives you some extra information as a bonus. We know not only the distance between heights, but the actual two points where they come closest to each other. The $\alpha$ and $\beta$ that we found tell us that point $Q$ divides $CM$ with ratio $18:17$ (almost in the middle), and point $R$ divides $DP$ with ratio $32:3$ (very close to point $P$).
