Step-by-step Solution:

1. Determine the direction cosines of the line.

Let the line make an angle $\alpha$ with the x-axis, $\beta$ with the y-axis, and $\gamma$ with the z-axis. The direction cosines are $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

The problem states that the line makes equal angles with the coordinate axes, so $\alpha = \beta = \gamma$. This implies that the direction cosines are equal: $l = m = n$.

We know the fundamental relation for direction cosines: $l^2 + m^2 + n^2 = 1$. Substituting $l=m=n$, we get: $l^2 + l^2 + l^2 = 1$ $3l^2 = 1$ $l^2 = \frac{1}{3}$ $l = \pm \frac{1}{\sqrt{3}}$ The problem also specifies that the line has positive direction cosines. Therefore, we must choose the positive value. $l = m = n = \frac{1}{\sqrt{3}}$

The direction ratios of the line can be taken as $(1, 1, 1)$, which are proportional to the direction cosines.

2. Write the equation of the line.

The line passes through the point P(2, -1, 2) and has direction ratios (1, 1, 1). The equation of the line in parametric form is: $\frac{x - 2}{1} = \frac{y - (-1)}{1} = \frac{z - 2}{1} = \lambda$ $\frac{x - 2}{1} = \frac{y + 1}{1} = \frac{z - 2}{1} = \lambda$ Any point on this line can be represented by the coordinates $(2 + \lambda, -1 + \lambda, 2 + \lambda)$ for some scalar $\lambda$.

3. Find the point of intersection Q.

The line meets the plane $2x + y + z = 9$ at point Q. This means the coordinates of Q must satisfy the plane's equation. Let the coordinates of Q be $(2 + \lambda, -1 + \lambda, 2 + \lambda)$.

Substituting these coordinates into the plane equation: $2(2 + \lambda) + (-1 + \lambda) + (2 + \lambda) = 9$ $4 + 2\lambda - 1 + \lambda + 2 + \lambda = 9$ $(4 - 1 + 2) + (2\lambda + \lambda + \lambda) = 9$ $5 + 4\lambda = 9$ $4\lambda = 9 - 5$ $4\lambda = 4$ $\lambda = 1$ Now, we can find the coordinates of point Q by substituting $\lambda = 1$ back into the parametric form: $x_Q = 2 + 1 = 3$ $y_Q = -1 + 1 = 0$ $z_Q = 2 + 1 = 3$ So, the point of intersection is Q(3, 0, 3).

4. Calculate the length of the line segment PQ.

We have the coordinates of P(2, -1, 2) and Q(3, 0, 3). We use the distance formula in 3D: $PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2}$ $PQ = \sqrt{(3 - 2)^2 + (0 - (-1))^2 + (3 - 2)^2}$ $PQ = \sqrt{(1)^2 + (1)^2 + (1)^2}$ $PQ = \sqrt{1 + 1 + 1}$ $PQ = \sqrt{3}$

Alternative Method:

The distance of the point $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$ from the point $(x_1, y_1, z_1)$ on the line $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$ is given by $|\lambda|\sqrt{a^2+b^2+c^2}$. In our case, $(a, b, c) = (1, 1, 1)$ and we found $\lambda=1$. So the distance PQ is: $PQ = |1| \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Both methods yield the same result. The length of the line segment PQ is $\sqrt{3}$.

Comparing this with the given options: A: $1$ B: ${\sqrt 2 }$ C: ${\sqrt 3 }$ D: $2$

The correct option is C.