$$\begin{gathered} \mathrm{If}{L}_1:\frac{x-a}{2}=\frac{y-2}{3}=\frac{z-b}{6},L_2:\frac{x-b}{3}=\frac{y-7}{6}=\frac{z-1}{3} \\ \mathrm{intersect}\mathrm{in}\mathrm{xy}\mathrm{plane}. \end{gathered}$$

$\mathrm{<br>}$

$\mathrm{Then},\mathrm{the}\mathrm{value}\mathrm{of}|a+b|\mathrm{is}$

The correct option is 15.

To find the value of $|a+b|$, we use the condition that the two lines intersect in the xy-plane.

Any point in the xy-plane has a z-coordinate equal to 0. Thus, at the point of intersection, we have:
$z=0$

Let the two lines be:
$L_1:\frac{x-a}{2}=\frac{y-2}{3}=\frac{z-b}{6}$
and
$L_2:\frac{x-b}{3}=\frac{y-7}{6}=\frac{z-1}{3}$

First, let's find the coordinates of the intersection point by setting $z=0$ in the equation of line $L_2$:
$\frac{x-b}{3}=\frac{y-7}{6}=\frac{0-1}{3}=-\frac{1}{3}$

From this relation, we can find the x and y coordinates of the intersection point:
1) Solving for x:
$\frac{x-b}{3}=-\frac{1}{3}\Rightarrow x-b=-1\Rightarrow x=b-1$
2) Solving for y:
$\frac{y-7}{6}=-\frac{1}{3}\Rightarrow y-7=-2\Rightarrow y=5$

Thus, the point of intersection is $(b-1,5,0)$.

Since this point also lies on the line $L_1$, it must satisfy its equation. Substituting $x=b-1$, $y=5$, and $z=0$ into the equation of $L_1$ gives:
$\frac{(b-1)-a}{2}=\frac{5-2}{3}=\frac{0-b}{6}$

Simplify the middle term:
$\frac{5-2}{3}=\frac{3}{3}=1$

So, the equations reduce to:
$\frac{b-a-1}{2}=1$
and
$-\frac{b}{6}=1$

From the second equation, we find:
$b=-6$

Substituting $b=-6$ into the first equation:
$\frac{-6-a-1}{2}=1\Rightarrow -a-7=2\Rightarrow -a=9\Rightarrow a=-9$

Now, we calculate the required value:
$|a+b|=|-9+(-6\left)\right|=|-15|=15$