If angle between the lines $\dfrac{x-1}{\propto} = \dfrac{y+2}{-5} = \dfrac{z+\frac{7}{3}}{\beta}$ and $x=z, y=0$ is $\dfrac{\pi}{4}$, then

We have two lines in 3D space, and the angle between them is $\dfrac{\pi}{4}$. We need to find the relationship between $\alpha$ and $\beta$.

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The first line is given in symmetric form:

$\dfrac{x-1}{\alpha} = \dfrac{y+2}{-5} = \dfrac{z+\frac{7}{3}}{\beta}$

so its direction ratios are simply the denominators:

$\vec{d_1} = (\alpha,\ -5,\ \beta)$

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The second line is given as $x = z,\ y = 0$. Setting $x = t$, we get $x = t,\ y = 0,\ z = t$, which gives:

$\vec{d_2} = (1,\ 0,\ 1)$

The coordinate $y$ is always $0$, meaning the line has no movement in the $y$-direction, so its direction ratio is $0$.

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The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:

$\cos\theta = \dfrac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$

We use the absolute value in the numerator because we always take the acute angle between two lines.

Here $\theta = \dfrac{\pi}{4}$, so $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$.

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Substituting the direction ratios:

Numerator:

$|\alpha \cdot 1 + (-5) \cdot 0 + \beta \cdot 1| = |\alpha + \beta|$

Denominator:

$\sqrt{\alpha^2 + 25 + \beta^2} \cdot \sqrt{1 + 0 + 1} = \sqrt{\alpha^2 + \beta^2 + 25} \cdot \sqrt{2}$

Setting up the equation:

$\dfrac{1}{\sqrt{2}} = \dfrac{|\alpha + \beta|}{\sqrt{2} \cdot \sqrt{\alpha^2 + \beta^2 + 25}}$

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Multiplying both sides by $\sqrt{2} \cdot \sqrt{\alpha^2 + \beta^2 + 25}$ and simplifying:

$\sqrt{\alpha^2 + \beta^2 + 25} = |\alpha + \beta|$

Squaring both sides:

$\alpha^2 + \beta^2 + 25 = \alpha^2 + 2\alpha\beta + \beta^2$

Cancelling $\alpha^2 + \beta^2$ from both sides:

$25 = 2\alpha\beta$

$\boxed{\alpha\beta = \dfrac{25}{2}}$