1. Interpret the equation

The given equation is $x(x+2)(12-k)=2.$

Treating this as a quadratic equation in $x$: $(12-k)x(x+2)=2$ $(12-k)(x^2+2x)-2=0$ $(12-k)x^2+2(12-k)x-2=0.$

So the quadratic is $a=(12-k),\quad b=2(12-k),\quad c=-2.$

2. Condition for equal roots

For equal roots, discriminant must be zero: $D=b^2-4ac=0.$

Substitute: $[2(12-k)]^2-4(12-k)(-2)=0$ $4(12-k)^2+8(12-k)=0$ $4(12-k)[(12-k)+2]=0$ $4(12-k)(14-k)=0.$

Thus, $12-k=0 \quad \text{or} \quad 14-k=0.$

So, $k=12 \quad \text{or} \quad k=14.$

But if $k=12$, then the original equation becomes $x(x+2)(0)=2 \Rightarrow 0=2,$ which is impossible.

Hence, $k=14.$

3. Find the point

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

4. Distance from the line

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

Here, line is $3x+4y+5=0,$ so $A=3$, $B=4$, $C=5$.

For point $(14,7)$: $d=\frac{|3(14)+4(7)+5|}{\sqrt{3^2+4^2}}$ $=\frac{|42+28+5|}{\sqrt{9+16}}$ $=\frac{75}{5}=15.$

5. Final answer

The required distance is $\boxed{15}.$

So the correct option is A.