The angle between the lines $L_1: \dfrac{x-5}{\alpha}=\dfrac{y+2}{-5}=\dfrac{5 z+24}{5 \beta}$ and $L_2: x=z, y=0$ is $\dfrac{\pi}{4}$, then the value of $(4 \alpha \beta)$ is

The line $L_1$ is given by $\dfrac{x-5}{\alpha} = \dfrac{y+2}{-5} = \dfrac{z + 24/5}{\beta}$. The direction vector of $L_1$ is $\vec{v_1} = \alpha \hat{i} - 5 \hat{j} + \beta \hat{k}$.

The line $L_2$ is given by $x=z$ and $y=0$. This can be written as $\dfrac{x}{1} = \dfrac{z}{1}$ with $y=0$. The direction vector of $L_2$ is $\vec{v_2} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k}$.

The angle $\theta$ between the lines is $\dfrac{\pi}{4}$. The cosine of the angle is given by $\cos \theta = \dfrac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.

Substituting the values, $\cos(\pi/4) = \dfrac{|\alpha(1) - 5(0) + \beta(1)|}{\sqrt{\alpha^2 + (-5)^2 + \beta^2} \sqrt{1^2 + 0^2 + 1^2}} = \dfrac{1}{\sqrt{2}}$.

This simplifies to $\dfrac{|\alpha + \beta|}{\sqrt{\alpha^2 + 25 + \beta^2} \sqrt{2}} = \dfrac{1}{\sqrt{2}}$, which implies $|\alpha + \beta| = \sqrt{\alpha^2 + 25 + \beta^2}$.

Squaring both sides, $(\alpha + \beta)^2 = \alpha^2 + 25 + \beta^2$, which gives $\alpha^2 + \beta^2 + 2\alpha\beta = \alpha^2 + 25 + \beta^2$.

Thus, $2\alpha\beta = 25$, so $4\alpha\beta = 50$.

Answer: 50