The equation of a line passing through origin with direction angles $\dfrac{2 \pi}{3}, \dfrac{\pi}{4}, \dfrac{\pi}{3}$ is

The direction angles of the line are given as $\alpha = \dfrac{2\pi}{3}$, $\beta = \dfrac{\pi}{4}$, and $\gamma = \dfrac{\pi}{3}$.

The direction cosines of the line are $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.

Calculating these values:
$l = \cos \left( \dfrac{2\pi}{3} \right) = -\dfrac{1}{2}$
$m = \cos \left( \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}}$
$n = \cos \left( \dfrac{\pi}{3} \right) = \dfrac{1}{2}$

The equation of a line passing through the origin $(0, 0, 0)$ with direction cosines $(l, m, n)$ is given by $\dfrac{x}{l} = \dfrac{y}{m} = \dfrac{z}{n}$.

Substituting the values:
$\dfrac{x}{-1/2} = \dfrac{y}{1/\sqrt{2}} = \dfrac{z}{1/2}$

Multiplying the denominators by $2$:
$\dfrac{x}{-1} = \dfrac{y}{\sqrt{2}} = \dfrac{z}{1}$

This can be rewritten as:
$-x = \dfrac{y}{\sqrt{2}} = z$

Multiplying by $-1$:
$x = \dfrac{y}{-\sqrt{2}} = -z$

Answer: $x=\dfrac{y}{-\sqrt{2}}=-z$