The given equations are $5x-3=15y+7=3-10z$.

First, we rewrite the equations in the standard form:

$\frac{5x-3}{1} = \frac{15y+7}{1} = \frac{3-10z}{1}$

Now, we divide each term to make the coefficients of $x, y, z$ equal to 1:

$\frac{5(x-\frac{3}{5})}{1} = \frac{15(y+\frac{7}{15})}{1} = \frac{-10(z-\frac{3}{10})}{1}$

$\frac{x-\frac{3}{5}}{\frac{1}{5}} = \frac{y+\frac{7}{15}}{\frac{1}{15}} = \frac{z-\frac{3}{10}}{-\frac{1}{10}}$

The direction ratios are $\frac{1}{5}, \frac{1}{15}, -\frac{1}{10}$.

To find the direction cosines, we first find the magnitude of the direction ratios:

$r = \sqrt{(\frac{1}{5})^2 + (\frac{1}{15})^2 + (-\frac{1}{10})^2} = \sqrt{\frac{1}{25} + \frac{1}{225} + \frac{1}{100}} = \sqrt{\frac{36+4+9}{900}} = \sqrt{\frac{49}{900}} = \frac{7}{30}$

The direction cosines are:

$l = \frac{1/5}{7/30} = \frac{1}{5} \cdot \frac{30}{7} = \frac{6}{7}$

$m = \frac{1/15}{7/30} = \frac{1}{15} \cdot \frac{30}{7} = \frac{2}{7}$

$n = \frac{-1/10}{7/30} = -\frac{1}{10} \cdot \frac{30}{7} = -\frac{3}{7}$

The direction cosines are $\frac{6}{7}, \frac{2}{7}, -\frac{3}{7}$.

A point through which the line passes is $(\frac{3}{5}, -\frac{7}{15}, \frac{3}{10})$.