The engineers must determine the shortest distance from the control station to the plane containing towers A, B and C in order to assess the efficiency of the communications links. The links will be considered reliable only if that distance is less than 8 km.

The coordinates are given as:
$A = (4, 6, 5)$
$B = (10, 8, 3)$
$D = (16, 10, 1)$

$\vec{AB} = B - A = (10-4, 8-6, 3-5) = (6, 2, -2)$
$\vec{AD} = D - A = (16-4, 10-6, 1-5) = (12, 4, -4)$

For A, B, and D to be collinear, $\vec{AD}$ must be a scalar multiple of $\vec{AB}$.
We check if $\vec{AD} = k \vec{AB}$ for some scalar $k$.

Comparing components:

Since $k=2$ for all components, the vectors are parallel and share a common point A, meaning A, B, and D are collinear.

Since tower D lies on the straight transmission path determined by towers A and B, the manager will approve its installation.
The manager will approve the installation of tower D.

$A = (4, 6, 5)$
$B = (10, 8, 3)$
$C = (6, 12, 1)$
$E = (4, 6, 10)$

The normal vector $\vec{n}$ is the cross product of $\vec{AB}$ and $\vec{AC}$.

We can use a simpler normal vector by dividing by 4: $\vec{n'} = (1, 5, 8)$.

Using point $A(4, 6, 5)$ and the normal vector $\vec{n'} = (1, 5, 8)$:

The distance $d$ from point $E(x_0, y_0, z_0) = (4, 6, 10)$ to the plane $Ax + By + Cz + D = 0$ is given by:

For the plane $x + 5y + 8z - 74 = 0$, we have $A=1, B=5, C=8, D=-74$.

Numerically, $d \approx \frac{4 \times 3.162277}{3} \approx 4.21637 \text{ km}$.

The links are considered reliable if the distance is less than 8 km.
Since $4.216 \text{ km} < 8 \text{ km}$, the communication links are reliable.

The shortest distance from the control station E to the plane containing towers A, B, and C is $\boxed{\frac{4\sqrt{10}}{3} \text{ km}}$.
The communication links are reliable.