The value of D is ..........

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

Lines $$L_1$$ and $$L_2$$ are
$$L_1:\; \sqrt{2}\,x+y-1=0,\qquad L_2:\; \sqrt{2}\,x-y+1=0.$$

The product of the distances of $$P$$ from these two lines is fixed: $$\lambda^{2}$$.
Hence $$\dfrac{\left|\sqrt{2}x+y-1\right|}{\sqrt{(\sqrt{2})^{2}+1^{2}}}\; \dfrac{\left|\sqrt{2}x-y+1\right|}{\sqrt{(\sqrt{2})^{2}+(-1)^{2}}} =\lambda^{2}.$$ Because $$\sqrt{(\sqrt{2})^{2}+1^{2}}=\sqrt{3},$$ the locus $$C$$ is

$$\bigl|\sqrt{2}x+y-1\bigr|\;\bigl|\sqrt{2}x-y+1\bigr|=3\lambda^{2}\equiv k.$$ Define $$k=3\lambda^{2}\;(\;k\gt 0\;).$$

Removing the absolute values by squaring gives a single equation $$(\sqrt{2}x+y-1)(\sqrt{2}x-y+1)=\pm k,$$ that is

$$2x^{2}-y^{2}+2y-1=\pm k \quad -(1).$$

Thus $$C$$ consists of the two quadratic curves
$$2x^{2}-y^{2}+2y-1=k \quad\text{and}\quad 2x^{2}-y^{2}+2y-1=-k.$$

Intersection of $$C$$ with the line $$y=2x+1$$
Substitute $$y=2x+1$$ into $$2x^{2}-y^{2}+2y-1$$:

$$\begin{aligned} 2x^{2}-(2x+1)^{2}+2(2x+1)-1 &=2x^{2}-\bigl(4x^{2}+4x+1\bigr)+4x+2-1\\ &=-2x^{2}. \end{aligned}$$

Therefore along the line $$y=2x+1$$ we have $$2x^{2}-y^{2}+2y-1=-2x^{2}.$$ Putting this in (1) gives

$$-2x^{2}=\pm k \;\Longrightarrow\; \begin{cases} -2x^{2}=k &\text{(impossible)},\\[4pt] 2x^{2}=k &\text{(possible)}. \end{cases}$$

Thus $$x^{2}=k/2$$ and the two intersection points are
$$R\bigl(\sqrt{k/2},\,2\sqrt{k/2}+1\bigr),\; S\bigl(-\sqrt{k/2},\,-2\sqrt{k/2}+1\bigr).$$

The squared distance $$RS^{2}$$ is

$$\begin{aligned} RS^{2}&=(2\sqrt{k/2})^{2}+(4\sqrt{k/2})^{2}\\ &=2k+8k\\ &=10k. \end{aligned}$$

Given that $$RS=\sqrt{270},$$ we have $$RS^{2}=270=10k,$$ so

$$k=27\quad -(2).$$

Equation of each branch of $$C$$
Using (2) in (1):
Branch-1: $$2x^{2}-y^{2}+2y-1=27,$$
Branch-2: $$2x^{2}-y^{2}+2y-1=-27.$$

Perpendicular bisector of $$RS$$
Segment $$RS$$ lies on $$y=2x+1$$ (slope $$2$$), so the perpendicular bisector has slope $$-\dfrac12$$.
Mid-point of $$R$$ and $$S$$ is $$M(0,1)$$, hence the bisector is

$$y-1=-\dfrac12(x-0)\;\;\Longrightarrow\;\;y=1-\dfrac{x}{2} \quad -(3).$$

Intersection of the bisector with $$C$$
Put $$y=1-\dfrac{x}{2}$$ into $$F(x,y)=2x^{2}-y^{2}+2y-1$$:

$$\begin{aligned} F(x,y)&=2x^{2}-\Bigl(1-\dfrac{x}{2}\Bigr)^{2}+2\Bigl(1-\dfrac{x}{2}\Bigr)-1\\ &=2x^{2}-\Bigl(1-x+\dfrac{x^{2}}{4}\Bigr)+2-\!x-1\\ &=\dfrac74\,x^{2}. \end{aligned}$$

Along the bisector $$F(x,y)=\dfrac74\,x^{2}.$$
For intersection with $$C$$ we need $$F=\pm k= \pm 27.$$ Since $$\dfrac74\,x^{2}\ge 0,$$ only the positive value works:

$$\dfrac74\,x^{2}=27\;\Longrightarrow\; x^{2}=\dfrac{108}{7},\; x=\pm6\sqrt{\dfrac{3}{7}}.$$ Corresponding $$y$$ values from (3) are
$$y=1-\dfrac{x}{2}: \begin{cases} R' \bigl(6\sqrt{\dfrac{3}{7}},\,1-3\sqrt{\dfrac{3}{7}}\bigr),\\[6pt] S' \bigl(-6\sqrt{\dfrac{3}{7}},\,1+3\sqrt{\dfrac{3}{7}}\bigr). \end{cases}$$

Distance $$R'S'$$
$$\begin{aligned} \Delta x &=12\sqrt{\dfrac{3}{7}},\quad \Delta y =6\sqrt{\dfrac{3}{7}},\\[4pt] R'S'^{2}&=(\Delta x)^{2}+(\Delta y)^{2}\\ &=144\cdot\dfrac{3}{7}+36\cdot\dfrac{3}{7}\\ &=\dfrac{432}{7}+\dfrac{108}{7}\\ &=\dfrac{540}{7}. \end{aligned}$$

Hence $$D=R'S'^{2}=\dfrac{540}{7}\approx77.14.$$ The required value lies in the range $$77.10-77.18.$$