The distance of the point ( 7 , − 2 , 11 ) ( 7 , − 2 , 11 ) from the line $\frac{x-<!--\; -\; -->6}{1}=\frac{y-<!--\; -\; -->4}{0}=\frac{z-<!--\; -\; -->8}{3}$ along the line $\frac{x-<!--\; -\; -->5}{2}=\frac{y-<!--\; -\; -->1}{-<!--\; -\; -->3}=\frac{z-<!--\; -\; -->5}{6}$ is :

The correct answer is 14.

To find the distance of the point P(7, -2, 11) from the line:
$L_1:\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$
measured along (or parallel to) the line:
$L_2:\frac{x-5}{2}=\frac{y-1}{��3}=\frac{z-5}{6}$
we will find the equation of a line passing through P(7, -2, 11) parallel to $L_2$ and determine its intersection point with $L_1$.

Step 1: Write the equation of the line passing through P and parallel to $L_2$
The direction ratios of the line $L_2$ are (2, -3, 6). Since the new line is parallel to $L_2$, it shares the same direction ratios.
Therefore, the equation of the line passing through P(7, -2, 11) in parametric form is:
$\frac{x-7}{2}=\frac{y+2}{-3}=\frac{z-11}{6}=\lambda$

Any general point Q on this line can be written in terms of $\lambda$ as:
$Q=(2\lambda +7,-3\lambda -2,6\lambda +11)$

Step 2: Find the point of intersection of this line with $L_1$
For Q to lie on the line $L_1$, its coordinates must satisfy the equation of $L_1$.
The equation of $L_1$ is:
$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$
Since the denominator of the y-term is 0, this implies that the y-coordinate is constant on this line:
$y-4=0\Rightarrow y=4$

Substituting the y-coordinate of Q:
$-3\lambda -2=4$
$-3\lambda =6$
$\lambda =-2$

Now, substitute $\lambda =-2$ back to find the coordinates of the intersection point Q:
$x=2(-2)+7=3$
$y=-3(-2)-2=4$
$z=6(-2)+11=-1$
So, the intersection point is Q(3, 4, -1).

Let us verify if Q(3, 4, -1) lies on $L_1$:
$\frac{3-6}{1}=-3$
$\frac{-1-8}{3}=-3$
Since $-3=-3$ and y = 4, Q lies on $L_1$.

Step 3: Calculate the distance between P and Q
Using the distance formula:
$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$
Substitute the coordinates of P(7, -2, 11) and Q(3, 4, -1):
$d=\sqrt{{(3-7)}^2+{(4-(-2\left)\right)}^2+{(-1-11)}^2}$
$d=\sqrt{{(-4)}^2+6^2+{(-12)}^2}$
$d=\sqrt{16+36+144}$
$d=\sqrt{196}$
$d=14$

The distance is 14.