Find the direction cosines of the line whose Cartesian equations are $\displaystyle 5 \mathrm{x}-3=15 \mathrm{y}+7=3-10 \mathrm{z}$.

Equations of given line can be written as $$\begin{aligned} & \frac{\mathrm{x}-\dfrac{3}{5}}{\dfrac{1}{5}}=\frac{\mathrm{y}+\dfrac{7}{15}}{\dfrac{1}{15}}=\frac{\mathrm{z}-\dfrac{3}{10}}{-\dfrac{1}{10}} \\ & \text { d.r.'s are }<\frac{1}{5}, \frac{1}{15},-\frac{1}{10}>\text { or }<6,2,-3> \\ & \text { d.c.'s are }<\frac{6}{7}, \frac{2}{7},-\frac{3}{7}> \end{aligned} $$