The plane passing through the line $L: l x-y+3(1-l) z=1, x+2 y-z=2$ and perpendicular to the plane $3 x+2 y+z=6$ is $3 x-8 y+7 z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis, then $415 \cos ^{2} \theta$ is equal to _____________.

<p>$L:lx - y + 3(1 - l)z = 1$, $x + 2y - z = 2$ and plane containing the line $p:3x - 8y + 7z = 4$</p> <p>Let $\overrightarrow n$ be the vector parallel to L.</p> <p>then $$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr l & { - 1} & {3(1 - l)} \cr 1 & 2 & { - 1} \cr } } \right|$$</p> <p>$= (6l - 5)\widehat i + (3 - 2l)\widehat j + (2l + 1)\widehat k$</p> <p>$\because$ R containing L</p> <p>$3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0$</p> <p>$18l + 16l + 14l - 15 - 24 + 7 = 0$</p> <p>$\therefore$ $l = {{32} \over {48}} = {2 \over 3}$</p> <p>Let $\theta$ be the acute angle between L and y-axis</p> <p>$\therefore$ $$\cos \theta = {{{5 \over 3}} \over {\sqrt {1 + {{25} \over 9} + {{49} \over 9}} }} = {5 \over {\sqrt {83} }}$$</p> <p>$\therefore$ $415{\cos ^2}\theta = 125$</p>