Given the expression for $w$ as :

$$w = z\bar{z} + k_1z + k_2iz + \lambda(1+i), \quad \text{where } k_1, k_2 \in \mathbb{R}.$$

1. If $w = x+iy$, we can separate this into the real and imaginary parts :

The real part is: $$\text{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0.$$

The imaginary part is: $$\text{Im}(w) = k_1y + k_2x + \lambda = 0.$$

2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$.

3. By comparing this with $\text{Re}(w) = 0$, we can see that $k_1=-2$, $k_2=4$, and $\lambda=4$.

4. The equation for the curve $\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$, which simplifies to $y = 2x + 2$.

5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$, with solutions $x = 0$ and $x = 2/5$.

6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$.

7. The distance between A and B is $AB = \sqrt{[(2/5)^2 + (4/5)^2]} = \sqrt{4/5}$.

8. Finally, $30(AB)^2 = 30 \times (4/5) = 24$.