Solve for x, y, and z in the following geometry problems.

The angles $x$, $2x$, and $3x$ lie on the straight line ABC. Angles on a straight line sum to $180^\circ$.

$$x + 2x + 3x = 180^\circ \quad (\text{Angles on a straight line})$$

$$x = \frac{180^\circ}{6}$$ $$x = 30^\circ$$

The value of $x$ is $\boxed{\text{30}^\circ}$.

The lines PQ and RS are parallel. The angle $y$ and the angle $30^\circ$ are alternate interior angles.

$$y = 30^\circ \quad (\text{Alternate interior angles, PQ } || \text{ RS})$$

The lines PQ and RS are parallel. The angle $z$ and the angle $42^\circ$ are alternate interior angles.

$$z = 42^\circ \quad (\text{Alternate interior angles, PQ } || \text{ RS})$$

Consider the triangle formed by the intersection of the two transversals and the line RS. The angles in this triangle are $y$, $x$, and $z$. The sum of angles in a triangle is $180^\circ$.

$$y + x + z = 180^\circ \quad (\text{Sum of angles in a triangle})$$

Substitute the values of $y$ and $z$:

$$30^\circ + x + 42^\circ = 180^\circ$$ $$72^\circ + x = 180^\circ$$ $$x = 180^\circ - 72^\circ$$ $$x = 108^\circ$$

The sizes are: $x = \boxed{\text{108}^\circ}$, $y = \boxed{\text{30}^\circ}$, $z = \boxed{\text{42}^\circ}$.

The angle $c$ and the angle $106^\circ$ are vertically opposite angles. Vertically opposite angles are equal.

$$c = 106^\circ \quad (\text{Vertically opposite angles})$$

The unknown value is $c = \boxed{\text{106}^\circ}$.