The distance of the point $(1,2)$ from the line $x+y+2=0$ measured along the line parallel to $2 x-y=5$ is equal to

The given point is $P(1, 2)$ and the line is $L: x + y + 2 = 0$.

The line passing through $P(1, 2)$ and parallel to $2x - y = 5$ has a slope $m = 2$.

The equation of this line is $y - 2 = 2(x - 1)$, which simplifies to $y = 2x$ or $2x - y = 0$.

To find the intersection point $Q$ of $x + y + 2 = 0$ and $2x - y = 0$, we substitute $y = 2x$ into the first equation:

$x + 2x + 2 = 0 \Rightarrow 3x = -2 \Rightarrow x = -\dfrac{2}{3}$.

Then $y = 2(-\dfrac{2}{3}) = -\dfrac{4}{3}$. So, $Q = (-\dfrac{2}{3}, -\dfrac{4}{3})$.

The distance $PQ$ is given by the distance formula:

$PQ = \sqrt{(1 - (-\dfrac{2}{3}))^2 + (2 - (-\dfrac{4}{3}))^2}$

$PQ = \sqrt{(\dfrac{5}{3})^2 + (\dfrac{10}{3})^2} = \sqrt{\dfrac{25}{9} + \dfrac{100}{9}} = \sqrt{\dfrac{125}{9}} = \dfrac{\sqrt{125}}{3} = \dfrac{5\sqrt{5}}{3}$.

Answer: $\dfrac{5 \sqrt{5}}{3}$