Given equations of the planes be
$3x-7y-5z-1=0..........(i)$
$8x-11y+2z=0........(ii)$
$5x-13y+3z+2=0..........(iii)$
Let the $d.r.$ of the line formed by $ (i)$ and $ (ii)$, $L_1$ be $(l_1,m_2,n_2)$ and
let the $d.r.$ of the line formed by $(ii)$ and $ (iii)$, $L_2$ be $(l_2,m_2,n_2)$
$\Rightarrow\:$ $(l_1,m_1,n_1)\:\:is\:\:\perp$ to $(3,-7,-5)$ and $(8,-11,2)$
$\Rightarrow\: (l_1,m_1,n_1)=(3,-7,-5)\times(8,-11,2)=(-69,-46,23)=(-3,-2,1)$
Similarly $(l_2,m_2,n_2)\:\:is\:\perp$ to $(5,-13,3)\:\:and\:\:(8,-11,2)$
$\Rightarrow\:(l_2,m_2,n_2)=(5,-13,3)\times(8,-11,2)=(7,14,49)=(1,2,7)$
$\therefore$ The angle between $L_1$ and $L_2$ is given by
$cos\theta=\large\frac{(l_1,m_1,n_!).(l_2,m_2,n_2)}{|(l_1,m_1,n_1)||(l_2,m_2,n_2)|}$
$=\large\frac{(-3,-2,1).(1,2,7)}{\sqrt {14}.\sqrt { 54}}$$=0$
$\therefore$ $\theta=\large\frac{\pi}{2}$