Let the equation $$x(x + 2)(12 - k) = 2$$ have equal roots. Then the distance of the point $$\left(k, \dfrac{k}{2}\right)$$ from the line $$3x + 4y + 5 = 0$$ is

Equation is $$x(x+2)(12-k)=2$$. For any fixed $$k$$, treat $$x$$ as the variable.

Divide both sides by $$(12-k)$$ (allowed if $$k \neq 12$$) to get a quadratic in $$x$$:

$$x(x+2)=\frac{2}{\,12-k\,}$$

Expand:

$$x^2+2x-\frac{2}{\,12-k\,}=0 \; -(1)$$

For equation $$(1)$$ to have equal (repeated) roots, its discriminant must be zero. For a quadratic $$ax^2+bx+c=0$$, discriminant $$\Delta=b^2-4ac$$.

Here $$a=1,\; b=2,\; c=-\dfrac{2}{\,12-k\,}$$. Set $$\Delta=0$$:

$$2^2-4(1)\!\left(-\frac{2}{\,12-k\,}\right)=0$$

$$4+\frac{8}{\,12-k\,}=0$$

Divide by $$4$$:

$$1+\frac{2}{\,12-k\,}=0$$

$$\frac{2}{\,12-k\,}=-1$$

Cross-multiply:

$$2=-(12-k)$$

$$k=14$$

Thus $$k=14$$ (note $$k\neq12$$ is satisfied).

The required point is $$\left(k,\frac{k}{2}\right)=\left(14,\,7\right)$$.

Distance of point $$(x_1,y_1)$$ from line $$Ax+By+C=0$$ is $$\displaystyle D=\frac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}}$$.

Here $$A=3,\; B=4,\; C=5,\; (x_1,y_1)=(14,7)$$:

Numerator: $$|3(14)+4(7)+5|=|42+28+5|=|75|=75$$

Denominator: $$\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=5$$

Therefore $$D=\frac{75}{5}=15$$

Hence the distance is $$15$$.

Option A is correct.