1. Interpret the two lines

The given line from which distance is to be measured is $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}.$

Since the direction ratios are $(1,0,3)$, a parametric form is $x=4+t,\quad y=4,\quad z=2+3t.$ So this line is $L_1:(4,4,2)+t(1,0,3).$

The distance is to be measured along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}.$ Its parametric form is $x=9+2\lambda,\quad y=13+3\lambda,\quad z=17+6\lambda.$ So this line is $L_2:(9,13,17)+\lambda(2,3,6).$

We are given the point $P=(7,10,11).$

2. Meaning of “distance of the point from the line along the line”

We must pass a line through $P$ parallel to $L_2$, and find where it meets $L_1$. The required distance is then the distance from $P$ to that intersection point, measured along that line.

So the line through $P$ parallel to $L_2$ is $\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\mu.$ Hence, $x=7+2\mu,\quad y=10+3\mu,\quad z=11+6\mu.$

3. Find intersection of this line with $L_1$

At intersection, coordinates satisfy both:

From $L_1$: $x=4+t,\quad y=4,\quad z=2+3t.$

From the line through $P$: $x=7+2\mu,\quad y=10+3\mu,\quad z=11+6\mu.$

Equating coordinates:

$4+t=7+2\mu \quad ...(1)$ $4=10+3\mu \quad ...(2)$ $2+3t=11+6\mu \quad ...(3)$

From (2): $3\mu=-6\implies \mu=-2.$

Substitute into (1): $4+t=7+2(-2)=3 \implies t=-1.$

Check in (3): $2+3(-1)=2-3=-1,$ $11+6(-2)=11-12=-1,$ which matches, so intersection exists.

Thus the intersection point is obtained from the line through $P$ at $\mu=-2$.

4. Compute the required distance

Along the line through $P$, direction vector is $\vec d=(2,3,6).$ Its magnitude is $|\vec d|=\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7.$

Since the parameter value is $\mu=-2$, the distance from $P$ to the intersection point is $|\mu|\,|\vec d|=2\times 7=14.$

5. Conclusion

The required distance is $\boxed{14}.$

So the correct option is D.