The distance of the point $(1,2)$ from the line $x+y=0$ measured parallel to the line $3 x-y=2$ is

The given point is $P = (1,2)$, with distance measured from line $L_1: x+y=0$ parallel to line $L_2: 3x-y=2$.

The direction vector for $L_2$ is derived from its slope $m=3$, giving the line through $P$ parallel to $L_2$ as $y - 2 = 3(x - 1)$, simplifying to $3x - y - 1 = 0$.

Intersecting this with $L_1$ by substituting $y = -x$ yields $3x - (-x) - 1 = 0$, so $4x = 1$ and $x = \frac{1}{4}$ with $y = -\frac{1}{4}$.

The distance between $(1,2)$ and $\left(\frac{1}{4}, -\frac{1}{4}\right)$ is found as $\sqrt{\left(\frac{-3}{4}\right)^2 + \left(\frac{-9}{4}\right)^2} = \sqrt{\frac{9}{16} + \frac{81}{16}} = \sqrt{\frac{90}{16}} = \frac{3\sqrt{10}}{4}$.

Final answer: $\frac{3\sqrt{10}}{4}$ units.