Given line $L_2 \equiv x+y-1=0$

Let a straight line passing through origin and having slope $m$ will be $L_1 \equiv y-mx = 0$

Given circle $x^2+y^2-x+3y=0$

Here centre $= C= \left(-g,-f\right) = \left(\frac{1}{2},\frac{-3}{2}\right) \& r = \sqrt{g^2+f^2 }=\raisebox{1ex}{$\sqrt{10}$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

Given that intercepts made by the lines are equal i.e both the lines will be equidistant from centre.

We know that perpendicular distance from $\left(x_1, y_1\right)$ to $ax+by+c=0$is $\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$.

$\Rightarrow \frac{\left|{\displaystyle -\frac{m}{2}}-{\displaystyle \frac{3}{2}}\right|}{\sqrt{1+m^2}}=\frac{\left|{\displaystyle \frac{1}{2}}-{\displaystyle \frac{3}{2}}-1\right|}{\sqrt{2}}$

Square on both sides

$\Rightarrow 7m^2-6m-1=0$

$\Rightarrow \left(7m+1\right)\left(m-1\right)=0$

$\Rightarrow m=\frac{-1}{7} , 1.$

So required line equations are $y = x \& y = \frac{-1}{7}x$

$\Rightarrow x-y = 0 \& x+7y=0$.