Concept:
Use the square of the magnitude of a vector, \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\). Equate the squared magnitudes to avoid square roots, then solve the resulting quadratic equation.
Step 1: Form the vectors \(\vec{a} - 2\vec{b}\) and \(2\vec{a} + \vec{b}\).
Given \(\vec{a} = x\hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}\):
\[
\vec{a} - 2\vec{b} = (x-4)\hat{i} + 4\hat{j} - 5\hat{k}, \quad 2\vec{a} + \vec{b} = (2x+2)\hat{i} + 3\hat{j} + 0\hat{k}.
\]
Step 2: Square the magnitudes and set them equal.
\[
|\vec{a} - 2\vec{b}|^2 = (x-4)^2 + 4^2 + (-5)^2 = (x-4)^2 + 41,
\]
\[
|2\vec{a} + \vec{b}|^2 = (2x+2)^2 + 3^2 + 0^2 = 4(x+1)^2 + 9.
\]
Equating: \((x-4)^2 + 41 = 4(x+1)^2 + 9\).
Step 3: Expand and simplify.
\[
x^2 - 8x + 16 + 41 = 4(x^2 + 2x + 1) + 9 \implies x^2 - 8x + 57 = 4x^2 + 8x + 13.
\]
\[
3x^2 + 16x - 44 = 0.
\]
Step 4: Factor and solve for positive \(x\).
\[
3x^2 + 22x - 6x - 44 = 0 \implies (3x+22)(x-2) = 0.
\]
Since \(x > 0\), \(x = 2\).
Step 5: Write the final answer.
\[
\boxed{2}
\]