If the two lines x + (a-1)y = 1 and 2x + a²y = 1 (a ∈ R - {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is:

(-1, a-1)(-2, a²) = -1(-2) = -(a²)(a-1)
a³ - a² + 2 = 0
(a+1)(a² - 2a + 2) = 0
Therefore, a = -1
The lines are x - 2y = 1 and 2x + y = 1
Solving these equations, we get x = 3/5 and y = -1/5
The point of intersection is (3/5, -1/5)
Distance from origin = √((3/5)² + (-1/5)²) = √(9/25 + 1/25) = √(10/25) = √(2/5)
Let's verify with a = -1
The lines are x - 2y = 1 and 2x + y = 1
Solving simultaneously:
2x + y = 1
2x - 4y = 2
Subtracting the two equations gives 5y = -1, so y = -1/5
Substituting into 2x + y = 1 gives 2x - 1/5 = 1, so 2x = 6/5 and x = 3/5
The point of intersection is (3/5, -1/5)
Distance from the origin is √((3/5)² + (-1/5)²) = √(10/25) = √(2/5) = 2√(5)/5
However, if the lines are perpendicular, then the product of their slopes is -1.
The slope of x + (a - 1)y = 1 is -1/(a - 1)
The slope of 2x + a²y = 1 is -2/a²
(-1/(a - 1))(-2/a²) = -1
2/(a³ - a²) = -1
2 = -a³ + a²
a³ - a² + 2 = 0
If a = -1, then (-1)³ - (-1)² + 2 = -1 - 1 + 2 = 0
So a = -1 is a solution.
If a = -1, the lines are x - 2y = 1 and 2x + y = 1
Solving these equations simultaneously gives x = 3/5 and y = -1/5
The point of intersection is (3/5, -1/5)
The distance from the origin is √((3/5)² + (-1/5)²) = √(10/25) = √(2/5)
This is not one of the options.