$The line which passes through the origin and intersects the two lines$ $\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{3},\frac{x-4}{2}=\frac{y+3}{3}=\frac{z-4}{4}is$

$\text{ If the lines }\overline{r}=\overline{a}+s\overline{b},r=\overline{c}+t\overline{d}\text{ intersect then }\left[\begin{array}{ccc}\overline{c}-\overline{a}& \overline{b}& \overline{d}\end{array}\right]=0$

$\text{ Let }\frac{x}{\alpha }=\frac{y}{\beta }=\frac{z}{\gamma }\text{ be the equation of required line }$

$\left|\begin{array}{llr}-1& 3& -5\\ 2& 4& 3\\ \alpha & \beta & \gamma \end{array}\right|=0\Rightarrow 29\alpha -7\beta -10\gamma =0 \to \left(1\right)$

$\left|\begin{array}{ccc}4& -3& 4\\ 2& 3& 4\\ \alpha & \beta & \gamma \end{array}\right|=0\Rightarrow 12\alpha +4\beta -9\gamma =0 \to \left(2\right)$

$\text{ From (1) and (2) }\frac{\alpha }{103}=\frac{\beta }{141}=\frac{\gamma }{200}$

$\therefore \text{ The line is}$ $\frac{x}{103}=\frac{y}{141}=\frac{z}{200}$