1. Write the two lines in slope form

The given lines are: $x+(a-1)y=1$ $2x+a^2y=1$

For a line $Ax+By=C$, slope is $m=-\frac{A}{B}$ (provided $B\neq 0$).

So the slopes are: $m_1=-\frac{1}{a-1}, \qquad m_2=-\frac{2}{a^2}$

2. Use the condition for perpendicular lines

If two lines are perpendicular, then $m_1m_2=-1$

Thus, $\left(-\frac{1}{a-1}\right)\left(-\frac{2}{a^2}\right)=-1$

$\frac{2}{a^2(a-1)}=-1$

$2=-a^2(a-1)$

$a^3-a^2+2=0$

3. Solve the cubic

We test simple roots. For $a=-1$, $(-1)^3-(-1)^2+2=-1-1+2=0$ So $(a+1)$ is a factor.

Divide: $a^3-a^2+2=(a+1)(a^2-2a+2)$

Now, $a^2-2a+2=(a-1)^2+1>0$ for all real $a$.

Hence the only real value is $a=-1$

4. Find the point of intersection

Substitute $a=-1$ into the lines:

First line: $x+(-2)y=1 \Rightarrow x-2y=1$

Second line: $2x+(-1)^2y=1 \Rightarrow 2x+y=1$

Solve the system: $x-2y=1$ $2x+y=1$

From the second equation, $y=1-2x$

Substitute into the first: $x-2(1-2x)=1$ $x-2+4x=1$ $5x=3$ $x=\frac{3}{5}$

Then, $y=1-2\cdot \frac{3}{5}=1-\frac{6}{5}=-\frac{1}{5}$

So the intersection point is $\left(\frac35,-\frac15\right)$

5. Distance from origin

Distance from origin to $\left(\frac35,-\frac15\right)$ is $d=\sqrt{\left(\frac35\right)^2+\left(-\frac15\right)^2}$ $d=\sqrt{\frac{9}{25}+\frac{1}{25}}$ $d=\sqrt{\frac{10}{25}}$ $d=\sqrt{\frac{2}{5}}$

6. Match with options

$\sqrt{\frac{2}{5}}$ corresponds to Option D.