Let a line L be perpendicular to both the lines L_1: (x+1)/(3) =...

The direction vector of line $L_1$ is $\vec{d_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$.

The direction vector of line $L_2$ is $\vec{d_2} = \hat{i} + 4\hat{j} + 7\hat{k}$.

Since line $L$ is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}$ is given by the cross product of $\vec{d_1}$ and $\vec{d_2}$:

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$

$\vec{d} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$

The direction ratios of $L$ are proportional to $1, -2, 1$.

The direction vector of line $L_3$ is $\vec{d_3} = 2\hat{i} + \hat{j} + 2\hat{k}$.

The acute angle $\theta$ between $L$ and $L_3$ is given by:

$\cos\theta = \dfrac{|\vec{d} \cdot \vec{d_3}|}{|\vec{d}| |\vec{d_3}|}$

$\cos\theta = \dfrac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}}$

$\cos\theta = \dfrac{|2 - 2 + 2|}{\sqrt{6} \sqrt{9}} = \dfrac{2}{3\sqrt{6}}$

From $\cos\theta = \dfrac{2}{3\sqrt{6}}$, considering a right-angled triangle with adjacent side $2$ and hypotenuse $3\sqrt{6}$:

The opposite side is $\sqrt{(3\sqrt{6})^2 - 2^2} = \sqrt{54 - 4} = \sqrt{50} = 5\sqrt{2}$.

Therefore, $\tan\theta = \dfrac{5\sqrt{2}}{2}$.

Answer: $\dfrac{5}{2}\sqrt{2}$