Let $$P$$ be a plane containing the line $$\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}$$ and parallel to the line $$\frac{x-3}{4} = \frac{y-2}{-3} = \frac{z+5}{7}$$. If the point $$(1, -1, \alpha)$$ lies on the plane $$P$$, then the value of $$|5\alpha|$$ is equal to ___.

The plane $$P$$ contains the line $$\dfrac{x-1}{3} = \dfrac{y+6}{4} = \dfrac{z+5}{2}$$ with direction vector $$\vec{d_1} = (3, 4, 2)$$ and is parallel to the line with direction vector $$\vec{d_2} = (4, -3, 7)$$. The normal to the plane is perpendicular to both directions, so $$\vec{n} = \vec{d_1} \times \vec{d_2}$$.

Computing the cross product: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = \hat{i}(28+6) - \hat{j}(21-8) + \hat{k}(-9-16) = (34, -13, -25)$$.

The first line passes through $$(1, -6, -5)$$, which lies on the plane. The equation of the plane is $$34(x-1) - 13(y+6) - 25(z+5) = 0$$, which simplifies to $$34x - 13y - 25z = 237$$.

Substituting the point $$(1, -1, \alpha)$$: $$34(1) - 13(-1) - 25\alpha = 237$$, giving $$47 - 25\alpha = 237$$, so $$\alpha = -\dfrac{190}{25} = -\dfrac{38}{5}$$.

Therefore $$|5\alpha| = \left|5 \cdot \left(-\dfrac{38}{5}\right)\right| = |-38| = 38$$.