The value of $$\lambda^2$$ is .........

The two given lines are
$$L_1 : \sqrt{2}\,x + y - 1 = 0$$
$$L_2 : \sqrt{2}\,x - y + 1 = 0$$

For a point $$P(x,y)$$, the distance from a line $$ax + by + c = 0$$ is $$\dfrac{|ax + by + c|}{\sqrt{a^2 + b^2}}$$.
Hence

Distance of $$P$$ from $$L_1$$ is$$\dfrac{\left|\sqrt{2}\,x + y - 1\right|}{\sqrt{3}}$$,    distance from $$L_2$$ is$$\dfrac{\left|\sqrt{2}\,x - y + 1\right|}{\sqrt{3}}$$.

The locus $$C$$ is defined by the condition
$$\dfrac{\left|\sqrt{2}\,x + y - 1\right|\;\left|\sqrt{2}\,x - y + 1\right|}{3}=\lambda^{2}$$ $$-(1)$$

The line $$y = 2x + 1$$ meets $$C$$ at the points $$R$$ and $$S$$. Substitute $$y = 2x + 1$$ in $$-(1)$$:

First factor: $$\sqrt{2}\,x + (2x + 1) - 1 = x(\sqrt{2} + 2)$$
Second factor: $$\sqrt{2}\,x - (2x + 1) + 1 = x(\sqrt{2} - 2)$$

Product of the two numerators:
$$x(\sqrt{2} + 2)\,x(\sqrt{2} - 2)=x^{2}\bigl((\sqrt{2})^{2} - 2^{2}\bigr)=x^{2}(2-4)=-2x^{2}$$

Thus $$-(1)$$ becomes $$\dfrac{\,|{-\,2x^{2}}|\,}{3}=\lambda^{2}\;\Longrightarrow\;\dfrac{2x^{2}}{3}=\lambda^{2}$$ or $$x^{2}=\dfrac{3}{2}\lambda^{2}$$ $$-(2)$$

Equation $$-(2)$$ gives two symmetric intersection abscissae $$x = \pm k,\ \text{where } k=\lambda\sqrt{\dfrac{3}{2}}$$

Coordinates of $$R$$ and $$S$$ are therefore $$R\bigl(k,\;2k+1\bigr),\qquad S\bigl(-k,\;-2k+1\bigr)$$

Distance $$RS$$ is given to be $$\sqrt{270}$$, so using the distance formula:

$$RS^{2}=(k-(-k))^{2}+\bigl[(2k+1)-(-2k+1)\bigr]^{2}\\[2pt] = (2k)^{2}+(4k)^{2}=4k^{2}+16k^{2}=20k^{2}$$

Since $$RS^{2}=270$$, $$20k^{2}=270\;\Longrightarrow\;k^{2}=\dfrac{27}{2}=13.5$$ $$-(3)$$

Substitute $$k^{2}$$ from $$-(3)$$ back in $$-(2)$$:

$$\lambda^{2}=\dfrac{2}{3}k^{2}=\dfrac{2}{3}\times\dfrac{27}{2}=9$$

Hence the required value is

$$\boxed{\,\lambda^{2}=9\,}$$

(The numerical range given in the paper, $$8.95-9.05$$, encloses this exact value.)