In a 3-dimensional coordinate system with $x$-, $y$-, and $z$-axes, $P$ is a point on the line $y = -x + 1$ in the $xy$-plane and $Q$ is a point on the line $z = -2x + 1$ in the $xz$-plane.  Compute the smallest possible value for $PQ.$
Let $P = (a, -a + 1, 0)$ be a point on the first line, and let $Q = (b, 0, -2b + 1)$ be a point on the second line.

[asy]
import three;

size(250);
currentprojection = perspective(6,3,2);

draw((-1,2,0)--(2,-1,0),red);
draw((3/2,0,-2)--(-1/2,0,2),blue);
draw((-2,0,0)--(2,0,0));
draw((0,-2,0)--(0,2,0));
draw((0,0,-2)--(0,0,2));

label("$x$", (2.2,0,0));
label("$y$", (0,2.2,0));
label("$z$", (0,0,2.2));
label("$y = -x + 1$", (-1,2,0), E, red);
label("$z = -2x + 1$", (3/2,0,-2), S, blue);
[/asy]

Then
\begin{align*}
PQ^2 &= (a - b)^2 + (-a + 1)^2 + (-2b + 1)^2 \\
&= 2a^2 - 2ab + 5b^2 - 2a - 4b + 2 \\
&= 2a^2 - (2b + 2) a + 5b^2 - 4b + 2.
\end{align*}If $b$ is fixed, then this quadratic in $a$ is minimized when $a = \frac{2b + 2}{4} = \frac{b + 1}{2}.$  Then
\begin{align*}
PQ^2 &= 2 \left( \frac{b + 1}{2} \right)^2 - (2b + 2) \cdot \frac{b + 1}{2} + 5b^2 - 4b + 2 \\
&= \frac{9}{2} b^2 - 5b + \frac{3}{2}.
\end{align*}This is minimized when $b = \frac{5}{9}.$  When $b = \frac{5}{9},$
\[PQ^2 = \frac{9}{2} \left( \frac{5}{9} \right)^2 - 5 \cdot \frac{5}{9} + \frac{3}{2} = \frac{1}{9},\]so the minimum value of $PQ$ is $\boxed{\frac{1}{3}}.$