The angle between the line of intersection of the two planes $\overline{\mathrm{r}}.\left(2\overline{\mathrm{i}}+2\overline{\mathrm{j}}-3\overline{\mathrm{k}}\right)=5,\overline{\mathrm{r}}.\left(3\overline{\mathrm{i}}+3\overline{\mathrm{j}}-5\overline{\mathrm{k}}\right)=3$

 and the line $\overline{\mathrm{r}}=3\overline{\mathrm{i}}+2\overline{\mathrm{j}}+\overline{\mathrm{k}}+\mathrm{t}\left(5\overline{\mathrm{i}}+5\overline{\mathrm{j}}-7\overline{\mathrm{k}}\right)$ is
 

The line of intersection of two planes $\overline{\mathrm{r}}.\left(2\overline{\mathrm{i}}+2\overline{\mathrm{j}}-3\overline{\mathrm{k}}\right)=5,\overline{\mathrm{r}}.\left(3\overline{\mathrm{i}}+3\overline{\mathrm{j}}-5\overline{\mathrm{k}}\right)=3$ is along the vector which is perpendicular to both the vectors  $2i+2j-3k,3i+3j-5k$

Hence, 

  $\begin{array}{rcl}p& =& \left|\begin{array}{ccc}i& j& k\\ 2& 2& -3\\ 3& 3& -5\end{array}\right|\\ & =& i\left(-1\right)-j\left(-1\right)+k\left(0\right)\\ & =& -i+j\end{array}$

The second line is $\overline{\mathrm{r}}=3\overline{\mathrm{i}}+2\overline{\mathrm{j}}+\overline{\mathrm{k}}+\mathrm{t}\left(5\overline{\mathrm{i}}+5\overline{\mathrm{j}}-7\overline{\mathrm{k}}\right)$

This is along the vector $q=5i+5j-7k$

The angle between two lines given is same as the angle  between two vectors $p,q$

Hence,

$\cos \theta =\frac{-5+5}{\sqrt{ }\sqrt{ }}=0$

Therefore, the angle between the given two lines is $\frac{\pi }{2}$