Consider the circle $C: x^2+y^2-6x-8y-11=0$. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle $x^2+y^2-\alpha x - \beta y - \gamma = 0$, then $\alpha + \beta + 2\gamma$ is equal to ________.

Let the foot of the perpendicular from the origin to the chord $AB$ be $P(h, k)$.

The slope of $OP$ is $\dfrac{k}{h}$. Since the chord $AB$ is perpendicular to $OP$, its equation is given by:

$y - k = -\dfrac{h}{k}(x - h) \Rightarrow hx + ky = h^2 + k^2$

This can be rewritten as $\dfrac{hx + ky}{h^2 + k^2} = 1$.

The chord $AB$ subtends a right angle at the origin. We homogenize the equation of the circle $x^2 + y^2 - 6x - 8y - 11 = 0$ with the equation of the chord to find the joint equation of the lines $OA$ and $OB$:

$x^2 + y^2 - (6x + 8y)\left(\dfrac{hx + ky}{h^2 + k^2}\right) - 11\left(\dfrac{hx + ky}{h^2 + k^2}\right)^2 = 0$

Since the lines $OA$ and $OB$ are perpendicular, the sum of the coefficients of $x^2$ and $y^2$ in this homogenized equation must be zero.

Coefficient of $x^2$: $1 - \dfrac{6h}{h^2 + k^2} - \dfrac{11h^2}{(h^2 + k^2)^2}$

Coefficient of $y^2$: $1 - \dfrac{8k}{h^2 + k^2} - \dfrac{11k^2}{(h^2 + k^2)^2}$

Equating the sum of these coefficients to zero:

$2 - \dfrac{6h + 8k}{h^2 + k^2} - \dfrac{11(h^2 + k^2)}{(h^2 + k^2)^2} = 0$

$2 - \dfrac{6h + 8k}{h^2 + k^2} - \dfrac{11}{h^2 + k^2} = 0$

$2(h^2 + k^2) - 6h - 8k - 11 = 0$

$h^2 + k^2 - 3h - 4k - \dfrac{11}{2} = 0$

Replacing $(h, k)$ with $(x, y)$, the locus of the foot of the perpendicular is:

$x^2 + y^2 - 3x - 4y - \dfrac{11}{2} = 0$

Comparing this with the given locus equation $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$, we get:

$\alpha = 3$, $\beta = 4$, $\gamma = \dfrac{11}{2}$

Therefore, the value of $\alpha + \beta + 2\gamma$ is:

$3 + 4 + 2\left(\dfrac{11}{2}\right) = 7 + 11 = 18$

Answer: $18$