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

$\begin{array}{l}{\overrightarrow{\mathrm{n}}}_1=\mathrm{\ell }\hat{\mathrm{i}}-\hat{\mathrm{j}}+3(1-\mathrm{\ell })\hat{\mathrm{k}}\\ {\overrightarrow{\mathrm{n}}}_2=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\end{array}$

Direction ratio of line$=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ \mathrm{\ell }& -1& 3(1-\mathrm{\ell })\\ 1& 2& -1\end{array}\right|$

$=(6\mathrm{\ell }-5)\hat{\mathrm{i}}+(3-2\mathrm{\ell })\hat{\mathrm{j}}+(2\mathrm{\ell }+1)\hat{\mathrm{k}}$

$3\mathrm{x}-8\mathrm{y}+7\mathrm{z}=4$ will contain the line $(6\mathrm{\ell }-5)\hat{\mathrm{i}}+(3-2\mathrm{\ell })\hat{\mathrm{j}}+(2\mathrm{\ell }+1)\hat{\mathrm{k}}$

Normal of 3x – 8y + 7z = 4 will be perpendicular to the line

$\begin{array}{l} =3(6\mathrm{\ell }-5)+(3-2\mathrm{\ell })(-8)+7(2\mathrm{\ell }+1)=0\\ \Rightarrow \mathrm{\ell }=\frac{2}{3}\end{array}$

$\therefore$ direction ratio of line $\left(-1, \frac{5}{3}, \frac{7}{3}\right)$

Angle with y axis

$\begin{array}{l}\cos \mathrm{\theta }=\frac{5/3}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}}\\ \cos \mathrm{\theta }=\frac{5}{\sqrt{83}}\\ \therefore 415{\cos }^2\mathrm{\theta }=\frac{25}{83}\times 415=125\end{array}$