The set of all real values of $$\lambda$$ for which exactly two common tangents can be drawn to the circles $$x^2 + y^2 - 4x - 4y + 6 = 0$$ and $$x^2 + y^2 - 10x - 10y + \lambda = 0$$ is the interval:

We start with the two given circles

$$x^{2}+y^{2}-4x-4y+6=0$$

and

$$x^{2}+y^{2}-10x-10y+\lambda =0.$$

For any circle written as $$x^{2}+y^{2}+2gx+2fy+c=0$$ the centre is $$(-g,\,-f)$$ and the radius is $$\sqrt{g^{2}+f^{2}-c}.$$

Comparing the first equation with the standard form, we have

$$2g_1=-4\;\; \Rightarrow\;\; g_1=-2,$$ $$2f_1=-4\;\; \Rightarrow\;\; f_1=-2,$$ $$c_1=6.$$

Hence the centre of the first circle is

$$C_1=(-g_1,\,-f_1)=(2,\,2),$$

and its radius is

$$r_1=\sqrt{g_1^{2}+f_1^{2}-c_1} =\sqrt{(-2)^{2}+(-2)^{2}-6} =\sqrt{4+4-6} =\sqrt{2}.$$

For the second circle, we get

$$2g_2=-10\;\; \Rightarrow\;\; g_2=-5,$$ $$2f_2=-10\;\; \Rightarrow\;\; f_2=-5,$$ $$c_2=\lambda.$$

Therefore the centre of the second circle is

$$C_2=(-g_2,\,-f_2)=(5,\,5),$$

and its radius is

$$r_2=\sqrt{g_2^{2}+f_2^{2}-c_2} =\sqrt{(-5)^{2}+(-5)^{2}-\lambda} =\sqrt{25+25-\lambda} =\sqrt{50-\lambda}.$$

The distance between the centres is

$$d=|C_1C_2| =\sqrt{(5-2)^{2}+(5-2)^{2}} =\sqrt{3^{2}+3^{2}} =\sqrt{9+9} =\sqrt{18} =3\sqrt{2}.$$

For two circles to have exactly two common tangents, they must intersect in two distinct points. The well-known condition for intersection is

$$|\,r_1-r_2\,| < d < r_1+r_2.$$

We now substitute $$r_1=\sqrt2,\; r_2=\sqrt{50-\lambda},\; d=3\sqrt2.$$

Right-hand inequality

$$d < r_1+r_2 \;\; \Longrightarrow \;\; 3\sqrt2 < \sqrt2+\sqrt{50-\lambda}.$$

Subtracting $$\sqrt2$$ from both sides gives

$$2\sqrt2 < \sqrt{50-\lambda}.$$

Left-hand inequality

Because $$\sqrt{50-\lambda} > \sqrt2$$ (deduced above), the absolute value sign can be removed, and we get

$$\sqrt{50-\lambda}-\sqrt2 < 3\sqrt2 \;\; \Longrightarrow \;\; \sqrt{50-\lambda} < 4\sqrt2.$$

Thus the combined inequality becomes

$$2\sqrt2 < \sqrt{50-\lambda} < 4\sqrt2.$$

We square each part to remove the square roots:

Left side:

$$(2\sqrt2)^2 = 8 \;\; \Longrightarrow \;\; 50-\lambda > 8,$$

so $$\lambda < 42.$$

Right side:

$$(4\sqrt2)^2 = 32 \;\; \Longrightarrow \;\; 50-\lambda < 32,$$

so $$\lambda > 18.$$

Combining the two results we get the open interval

$$18 < \lambda < 42.$$

The radius $$r_2=\sqrt{50-\lambda}$$ is real as long as $$50-\lambda \ge 0,$$ i.e. $$\lambda \le 50,$$ which is automatically satisfied by every $$\lambda$$ in the interval above. Hence no further restriction arises.

Therefore, for $$\lambda$$ lying strictly between 18 and 42, the two circles intersect in two points and possess exactly two common tangents.

Hence, the correct answer is Option B.