For the linear programming problem, $x + 2y \leq 10,\ 3x + y \leq 12,\ x, y \geq 0$, the maximum value of $z = 5x + 10y$ occurs at every point on the line segment joining the points..

The given constraints are:
$x + 2y \leq 10$
$3x + y \leq 12$
$x, y \geq 0$

To find the corner points of the feasible region, we solve the equations of the boundary lines:
$x + 2y = 10$
$3x + y = 12$

Multiplying the second equation by 2, we get $6x + 2y = 24$. Subtracting the first equation from this gives:
$5x = 14 \Rightarrow x = \dfrac{14}{5}$

Substituting $x$ back into the first equation:
$\dfrac{14}{5} + 2y = 10 \Rightarrow 2y = \dfrac{36}{5} \Rightarrow y = \dfrac{18}{5}$

The intersection point is $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$.

The corner points of the feasible region are $(0, 0)$, $(4, 0)$, $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$, and $(0, 5)$.

Now, we evaluate the objective function $z = 5x + 10y$ at each corner point:
At $(0, 0)$: $z = 5(0) + 10(0) = 0$
At $(4, 0)$: $z = 5(4) + 10(0) = 20$
At $(0, 5)$: $z = 5(0) + 10(5) = 50$
At $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$: $z = 5\left(\dfrac{14}{5}\right) + 10\left(\dfrac{18}{5}\right) = 14 + 36 = 50$

The maximum value of $z$ is $50$, which occurs at two adjacent corner points: $(0, 5)$ and $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$.

Since the maximum value is the same at these two points, it occurs at every point on the line segment joining $(0, 5)$ and $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$.

Answer: $(0,5)$ and $\left(\dfrac{14}{5}, \dfrac{18}{5}\right)$