1. Find the centre and radius of the circle

Given circle: $x^2+y^2-2x-4y-11=0$

Complete the squares: $x^2-2x+y^2-4y=11$ $(x-1)^2-1+(y-2)^2-4=11$ $(x-1)^2+(y-2)^2=16$

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

2. Use the condition that the ellipse passes through the centre of the circle

Ellipse: $3x^2+py^2=4$

Since it passes through $C=(1,2)$, $3(1)^2+p(2)^2=4$ $3+4p=4$ $4p=1$ $p=\frac14.$

Hence the ellipse is $3x^2+\frac14 y^2=4.$

Multiply by $4$: $12x^2+y^2=16$

Write in standard form: $\frac{x^2}{4/3}+\frac{y^2}{16}=1.$

So $a^2=16,\quad b^2=\frac43$ with major axis along the $y$-axis.

3. Find the foci of the ellipse

For the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$ we have $c^2=a^2-b^2=16-\frac43=\frac{44}{3}.$ Thus $c=\sqrt{\frac{44}{3}}.$

So the foci are $F_1=(0,c),\qquad F_2=(0,-c).$

4. Find the focal distances of the point $C=(1,2)$

Let $f_1=CF_1,\qquad f_2=CF_2.$

Then $f_1^2=(1-0)^2+(2-c)^2=1+4-4c+c^2=5+c^2-4c,$ $f_2^2=(1-0)^2+(2+c)^2=1+4+4c+c^2=5+c^2+4c.$

Therefore, $f_1f_2=\sqrt{(5+c^2-4c)(5+c^2+4c)}$ $=\sqrt{(5+c^2)^2-(4c)^2}.$

Since $c^2=\frac{44}{3},$ we get $5+c^2=5+\frac{44}{3}=\frac{59}{3}.$

Hence $f_1f_2=\sqrt{\left(\frac{59}{3}\right)^2-16\cdot \frac{44}{3}}$ $=\sqrt{\frac{3481}{9}-\frac{704}{3}}$ $=\sqrt{\frac{3481-2112}{9}}$ $=\sqrt{\frac{1369}{9}}$ $=\frac{37}{3}.$

5. Compute the required quantity

$6f_1f_2-r=6\left(\frac{37}{3}\right)-4$ $=74-4=70.$

6. Match with the options

$70$ corresponds to Option C.

So the correct answer is C.