Let the ellipse $$3x^2 + py^2 = 4$$ pass through the centre C of the circle $$x^2 + y^2 - 2x - 4y - 11 = 0$$ of radius r. Let $$f_1$$, $$f_2$$ be the focal distances of the point C on the ellipse. Then $$6f_1f_2 - r$$ is equal to

The centre of the circle $$x^{2}+y^{2}-2x-4y-11=0$$ is obtained by completing the squares.
$$x^{2}-2x+1+y^{2}-4y+4=11+1+4$$
$$\Rightarrow (x-1)^{2}+(y-2)^{2}=16$$

Hence the centre is $$C(1,2)$$ and the radius is $$r=4$$.

The ellipse is $$3x^{2}+py^{2}=4$$ and it passes through the point $$C(1,2)$$, so

$$3(1)^{2}+p(2)^{2}=4$$
$$\Rightarrow 3+4p=4$$
$$\Rightarrow 4p=1 \;\Longrightarrow\; p=\frac14$$

Therefore the ellipse is $$3x^{2}+\frac14\,y^{2}=4$$.

Divide by $$4$$ to get the standard form:

$$\frac{3}{4}x^{2}+\frac{1}{16}y^{2}=1$$
$$\Rightarrow \frac{x^{2}}{\frac{4}{3}}+\frac{y^{2}}{16}=1$$

Comparing with $$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$$ (major axis along $$y$$-axis), we have

$$a^{2}=16,\;a=4,\qquad b^{2}=\frac{4}{3}$$

The focal length satisfies $$c^{2}=a^{2}-b^{2}$$, hence

$$c^{2}=16-\frac{4}{3}=\frac{48-4}{3}=\frac{44}{3},\qquad c=\sqrt{\frac{44}{3}}$$

The foci are $$F_{1}(0,c)$$ and $$F_{2}(0,-c)$$.

Let the focal distances of the point $$C(1,2)$$ be $$f_{1}=CF_{1}$$ and $$f_{2}=CF_{2}$$.

Since $$C$$ lies on the ellipse, the sum of its focal distances equals the major axis length:

$$f_{1}+f_{2}=2a=8 \quad -(1)$$

Now compute the squares of the distances.

$$f_{1}^{2}=(1-0)^{2}+(2-c)^{2}=1+(2-c)^{2} =1+4-4c+c^{2}=5-4c+c^{2}$$
$$f_{2}^{2}=(1-0)^{2}+(2+c)^{2}=1+(2+c)^{2} =1+4+4c+c^{2}=5+4c+c^{2}$$

Adding:

$$f_{1}^{2}+f_{2}^{2}=(5-4c+c^{2})+(5+4c+c^{2}) =10+2c^{2} \quad -(2)$$

From identities,

$$(f_{1}+f_{2})^{2}=f_{1}^{2}+f_{2}^{2}+2f_{1}f_{2} \quad -(3)$$

Substitute $$-(1)$$ and $$-(2)$$ into $$-(3)$$:

$$8^{2}=10+2c^{2}+2f_{1}f_{2}$$
$$64=10+2\left(\frac{44}{3}\right)+2f_{1}f_{2}$$
$$64=10+\frac{88}{3}+2f_{1}f_{2}$$
$$64=\frac{30}{3}+\frac{88}{3}+2f_{1}f_{2} =\frac{118}{3}+2f_{1}f_{2}$$

$$\Rightarrow 2f_{1}f_{2}=64-\frac{118}{3} =\frac{192-118}{3}=\frac{74}{3}$$

$$\therefore\; f_{1}f_{2}=\frac{37}{3}$$

Finally,

$$6f_{1}f_{2}-r=6\left(\frac{37}{3}\right)-4 =2\cdot37-4 =74-4 =70$$

Hence $$6f_{1}f_{2}-r=70$$, which matches Option C.