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

The line $$L$$ is the intersection of the planes

$$lx-y+3(1-l)z-1=0$$ and $$x+2y-z-2=0$$

Any plane passing through $$L$$ is

$$\left(lx-y+3(1-l)z-1\right)+\lambda(x+2y-z-2)=0$$

$$=(l+\lambda)x+(2\lambda-1)y+(3-3l-\lambda)z-(1+2\lambda)=0$$

Given plane is

$$3x-8y+7z-4=0$$

Comparing coefficients,

$$\frac{l+\lambda}{3}=\frac{2\lambda-1}{-8}=\frac{3-3l-\lambda}{7}=\frac{1+2\lambda}{4}$$

Using

$$\frac{2\lambda-1}{-8}=\frac{1+2\lambda}{4},$$

$$4(2\lambda-1)=-8(1+2\lambda)$$

$$8\lambda-4=-8-16\lambda$$

$$24\lambda=-4$$

$$\lambda=-\frac16$$

Now,

$$\frac{l-\frac16}{3}=\frac{1+2\left(-\frac16\right)}{4}$$

$$\frac{l-\frac16}{3}=\frac{2/3}{4}=\frac16$$

$$l-\frac16=\frac12$$

$$l=\frac23$$

Hence, normals of the two planes are

$$\vec n_1=(2,-3,3)$$

and

$$\vec n_2=(1,2,-1)$$

Therefore, direction vector of the line is

$$\vec d=\vec n_1\times\vec n_2$$

$$= \begin{vmatrix} \hat i&\hat j&\hat k\\ 2&-3&3\\ 1&2&-1 \end{vmatrix}$$

$$=\hat i(3-6)-\hat j(-2-3)+\hat k(4+3)$$

$$=-3\hat i+5\hat j+7\hat k$$

Thus,

$$\vec d=(-3,5,7)$$

Let $$\theta$$ be the acute angle between the line and the $$y$$-axis.

Then,

$$\cos\theta=\frac{|5|}{\sqrt{(-3)^2+5^2+7^2}}$$

$$=\frac5{\sqrt{9+25+49}}$$

$$=\frac5{\sqrt{83}}$$

Therefore,

$$\cos^2\theta=\frac{25}{83}$$

Hence,

$$415\cos^2\theta=415\cdot\frac{25}{83}$$

$$=5\cdot25$$

$$=125$$

Therefore, the required value is

$$\boxed{125}$$.