If the two lines $x+\left(a-1\right)y=1$ and $2x+a^{2}y=1,\left(a\in R-\left\{\mathrm{0,1}\right\}\right)$ are perpendicular, then the distance of their point of intersection from the origin is

If two lines $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$ are perpendicular, then $a_1{a}_2+b_1{b}_2=0.$

Since, given lines $x+\left(a-1\right)y=1$ and $2x+a^{2}y=1$ are perpendicular, hence, we get

$1\cdot 2+\left(a-1\right)\cdot a^2=0$

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

$\Rightarrow \left(a+1\right)\cdot \left(a^2-2a+2\right)=0$

$\Rightarrow a=-1$ or $a^2-2a+2=0$

But, the discriminant of the quadratic equation is ${\left(-2\right)}^2-4\times 1\times 2=4-8=-4,$ hence, no real roots exist.

Thus, the only possible value of $a$ is $-1.$

And, hence the lines are $x-2y=1$ and $2x+y=1.$

Point of intersection of these two lines is $\left(\frac{3}{5}, -\frac{1}{5}\right)$

We know that, the distance of a point $\left(x, y\right)$ from origin is $\sqrt{x^2+y^2}$
Hence, distance of the point $\left(\frac{3}{5}, -\frac{1}{5}\right)$ from origin $=\sqrt{\frac{9}{25}+\frac{1}{25}}$$=\sqrt{\frac{2}{5}}.$