Let a line L be perpendicular to both the lines
$$L_1: \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$$ and $$L_2: \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$$.
If $$\theta$$ is the acute angle between the lines L and
$$L_3: \frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}$$, then $$\tan\theta$$ is equal to:

The direction vectors of $$L_1$$ and $$L_2$$ are:

$$\vec{d}_1 = 3\hat{i} + 5\hat{j} + 7\hat{k}$$

$$\vec{d}_2 = \hat{i} + 4\hat{j} + 7\hat{k}$$

Since $$L$$ is perpendicular to both $$L_1$$ and $$L_2$$, its direction vector $$\vec{d}$$ is:

$$\vec{d} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$$

$$\vec{d} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5)$$

$$\vec{d} = 7\hat{i} - 14\hat{j} + 7\hat{k}$$ or $$\vec{u} = \hat{i} - 2\hat{j} + \hat{k}$$

$$L_3: \frac{x - \frac{7}{8}}{2} = \frac{y - \frac{7}{4}}{1} = \frac{z}{2}$$

$$\vec{v} = 2\hat{i} + \hat{j} + 2\hat{k}$$

$$\vec{u} \cdot \vec{v} = (1)(2) + (-2)(1) + (1)(2) = 2 - 2 + 2 = 2$$

$$|\vec{u}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$$

$$|\vec{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3$$

$$\cos\theta = \frac{2}{3\sqrt{6}}$$

$$\text{perpendicular} = \sqrt{\text{hypotenuse}^2 - \text{base}^2} = \sqrt{(3\sqrt{6})^2 - 2^2}$$

$$\text{perpendicular} = \sqrt{54 - 4} = \sqrt{50} = 5\sqrt{2}$$

$$\tan\theta = \frac{\text{perpendicular}}{\text{base}} = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}$$