Let $$L_{1}: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$$ and $$L_{2}: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$$ be two lines. Let $$L_{3}$$ be a line passing through the point $$(\alpha ,\beta ,\gamma)$$ and be perpendicular to both $$L_{1}$$ and $$L_{2}$$. If $$L_{3}$$ intersects $$L_{1}$$, then $$|5\alpha -11\beta -8\gamma|$$ equals:

The direction vectors of the given lines are:

For $$L_1$$: $$\vec{d_1} = \langle 1, -1, 2 \rangle$$

For $$L_2$$: $$\vec{d_2} = \langle -1, 2, 1 \rangle$$

The direction vector of $$L_3$$, perpendicular to both, is the cross product:

$$\vec{d_3} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \hat{i}((-1)(1) - (2)(2)) - \hat{j}((1)(1) - (2)(-1)) + \hat{k}((1)(2) - (-1)(-1)) = \hat{i}(-1 - 4) - \hat{j}(1 + 2) + \hat{k}(2 - 1) = \langle -5, -3, 1 \rangle$$

Thus, the direction ratios of $$L_3$$ are $$-5, -3, 1$$.

Parametric equations for $$L_3$$ passing through $$(\alpha, \beta, \gamma)$$ are:

$$x = \alpha - 5t$$
$$y = \beta - 3t$$
$$z = \gamma + t$$

Parametric equations for $$L_1$$ are:

$$x = 1 + s$$
$$y = 2 - s$$
$$z = 1 + 2s$$

Since $$L_3$$ intersects $$L_1$$, set the coordinates equal:

$$1 + s = \alpha - 5t$$   $$-(1)$$
$$2 - s = \beta - 3t$$   $$-(2)$$
$$1 + 2s = \gamma + t$$   $$-(3)$$

Adding equations (1) and (2):

$$(1 + s) + (2 - s) = \alpha - 5t + \beta - 3t$$
$$3 = \alpha + \beta - 8t$$
$$\alpha + \beta - 8t = 3$$   $$-(4)$$

Solving equation (1) for $$s$$:

$$s = \alpha - 5t - 1$$

Substituting into equation (3):

$$1 + 2(\alpha - 5t - 1) = \gamma + t$$
$$1 + 2\alpha - 10t - 2 = \gamma + t$$
$$2\alpha - 10t - 1 = \gamma + t$$
$$2\alpha - \gamma - 11t = 1$$   $$-(5)$$

Eliminating $$t$$ by multiplying equation (4) by 11 and equation (5) by 8:

$$11(\alpha + \beta - 8t) = 11 \times 3$$   ⇒   $$11\alpha + 11\beta - 88t = 33$$   $$-(6)$$
$$8(2\alpha - \gamma - 11t) = 8 \times 1$$   ⇒   $$16\alpha - 8\gamma - 88t = 8$$   $$-(7)$$

Subtracting equation (7) from equation (6):

$$(11\alpha + 11\beta - 88t) - (16\alpha - 8\gamma - 88t) = 33 - 8$$
$$11\alpha + 11\beta - 88t - 16\alpha + 8\gamma + 88t = 25$$
$$-5\alpha + 11\beta + 8\gamma = 25$$

Rearranging:

$$5\alpha - 11\beta - 8\gamma = -25$$

Taking absolute value:

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

The value is 25.