Step 1. Identify the Given Lines

The line

$$ \mathrm{L}_1:\; \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2} $$

passes through the point

$$ P_1=(1,2,1) $$

with direction vector

$$ \mathbf{u}=(1,-1,2). $$

Similarly, the line

$$ \mathrm{L}_2:\; \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} $$

passes through the point

$$ P_2=(-1,2,0) $$

with direction vector

$$ \mathbf{v}=(-1,2,1). $$

Step 2. Determine the Direction of $ L_3 $

Since $ L_3 $ is perpendicular to both $ L_1 $ and $ L_2 $, its direction vector must be parallel to the cross product of $ \mathbf{u} $ and $ \mathbf{v} $. Compute:

$$ \mathbf{d} = \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}. $$

Using the determinant, we obtain:

$$ \begin{aligned} d_x &= (-1)(1) - (2)(2) = -1-4=-5,\\[1mm] d_y &= (2)(-1) - (1)(1) = -2-1=-3,\\[1mm] d_z &= (1)(2) - (-1)(-1) = 2-1=1. \end{aligned} $$

Thus, a valid direction vector is

$$ \mathbf{d}=(-5,-3,1). $$

Step 3. Express $ L_3 $ in Terms of Its Intersection with $ L_1 $

Since $ L_3 $ intersects $ L_1 $, let the intersection point (on $ L_1 $) be expressed using a parameter $ t $ as:

$$ A(t) = (1+t,\, 2-t,\, 1+2t). $$

Since $ L_3 $ passes through $ A(t) $ and also through the given point

$$ Q=(\alpha,\beta,\gamma), $$

its parametric form can be written as:

$$ (\alpha,\beta,\gamma)=A(t) + s\,\mathbf{d} = (1+t,\,2-t,\,1+2t) + s(-5,-3,1) $$

for some parameters $ s $ and $ t $.

Step 4. Express $\alpha,\beta,\gamma$ in Terms of $t$ and $s$

Comparing coordinates, we have:

$$ \begin{cases} \alpha = 1 + t - 5s,\\[1mm] \beta = 2 - t - 3s,\\[1mm] \gamma = 1 + 2t + s. \end{cases} $$

Step 5. Compute $5\alpha - 11\beta - 8\gamma$

Substitute the expressions for $\alpha$, $\beta$, and $\gamma$:

$$ \begin{aligned} 5\alpha - 11\beta - 8\gamma &= 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)\\[1mm] &= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)\\[1mm] &= \left(5 - 22 - 8\right) + \left(5t + 11t - 16t\right) + \left(-25s + 33s - 8s\right)\\[1mm] &= -25 + 0t + 0s\\[1mm] &= -25. \end{aligned} $$

Taking the absolute value yields:

$$ \left|5\alpha-11\beta-8\gamma\right| = 25. $$

Conclusion

The value of

$$ \left|5\,\alpha - 11\,\beta - 8\,\gamma\right| $$ is $$ \boxed{25}. $$