Distance Between Point and Plane Intersection Along Line

To find the length of the line segment $PQ$, let the direction cosines of the line be $l, m, n$.

Since the line makes equal angles with the coordinate axes and has positive direction cosines, we have:
$l = m = n > 0$

Using the fundamental relation between direction cosines, $l^2 + m^2 + n^2 = 1$, we obtain:
$3l^2 = 1 \implies l = m = n = \frac{1}{\sqrt{3}}$

The equation of the line passing through the point $P(2, -1, 2)$ in parametric form with distance parameter $r$ is:
$\frac{x - 2}{1/\sqrt{3}} = \frac{y + 1}{1/\sqrt{3}} = \frac{z - 2}{1/\sqrt{3}} = r$

Thus, the coordinates of any point on the line at a distance $r$ from $P$ can be expressed as:
$x = 2 + \frac{r}{\sqrt{3}}, \quad y = -1 + \frac{r}{\sqrt{3}}, \quad z = 2 + \frac{r}{\sqrt{3}}$

Since the line meets the plane $2x + y + z = 9$ at the point $Q$, these coordinates must satisfy the equation of the plane:
$2\left(2 + \frac{r}{\sqrt{3}}\right) + \left(-1 + \frac{r}{\sqrt{3}}\right) + \left(2 + \frac{r}{\sqrt{3}}\right) = 9$

Simplifying the equation:
$(4 - 1 + 2) + \frac{r}{\sqrt{3}}(2 + 1 + 1) = 9$
$5 + \frac{4r}{\sqrt{3}} = 9$
$\frac{4r}{\sqrt{3}} = 4 \implies r = \sqrt{3}$

The distance parameter $r$ directly represents the length of the segment $PQ$. Thus:
$PQ = |r| = \sqrt{3}$

Alternatively, the coordinates of $Q$ are:
$Q = \left(2 + 1, -1 + 1, 2 + 1\right) = (3, 0, 3)$

Using the distance formula:
$PQ = \sqrt{(3 - 2)^2 + (0 - (-1))^2 + (3 - 2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Therefore, the correct option is C.