The plane passing through the line $$L:l 𝑥 - 𝑦 + 31 - 𝑙 𝑧 = 1, 𝑥 + 2𝑦 - 𝑧 = 2$$ 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 defined by the intersection of two planes:

Plane 1: $$l x - y + (31 - l) z = 1$$, with normal vector $$\vec{n_1} = (l, -1, 31 - l)$$

Plane 2: $$x + 2y - z = 2$$, with normal vector $$\vec{n_2} = (1, 2, -1)$$

The direction vector $$\vec{d}$$ of line $$L$$ is given by the cross product $$\vec{n_1} \times \vec{n_2}$$:

$$$\vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ l & -1 & 31 - l \\ 1 & 2 & -1 \\ \end{vmatrix} = \vec{i}((-1)(-1) - (31 - l)(2)) - \vec{j}((l)(-1) - (31 - l)(1)) + \vec{k}((l)(2) - (-1)(1))$$$

Simplifying each component:

$$\vec{i}$$-component: $$1 - 2(31 - l) = 1 - 62 + 2l = 2l - 61$$

$$\vec{j}$$-component: $$-\left(-l - (31 - l)\right) = -\left(-31\right) = 31$$

$$\vec{k}$$-component: $$2l - (-1) = 2l + 1$$

Thus, $$\vec{d} = (2l - 61, 31, 2l + 1)$$.

The plane $$3x - 8y + 7z = 4$$ passes through line $$L$$ and is perpendicular to the plane $$3x + 2y + z = 6$$. The normal vector of the given perpendicular plane is $$\vec{n} = (3, -8, 7)$$. Since the plane contains line $$L$$, the direction vector $$\vec{d}$$ must be perpendicular to $$\vec{n}$$, so $$\vec{d} \cdot \vec{n} = 0$$:

$$$(2l - 61)(3) + (31)(-8) + (2l + 1)(7) = 0$$$

Expanding:

$$$6l - 183 - 248 + 14l + 7 = 0$$$

Combining like terms:

$$$20l - 424 = 0$$$

Solving for $$l$$:

$$$20l = 424 \implies l = \frac{424}{20} = \frac{106}{5}$$$

Substitute $$l = \frac{106}{5}$$ into $$\vec{d}$$:

$$$2l - 61 = 2\left(\frac{106}{5}\right) - 61 = \frac{212}{5} - \frac{305}{5} = -\frac{93}{5}$$$

$$$2l + 1 = 2\left(\frac{106}{5}\right) + 1 = \frac{212}{5} + \frac{5}{5} = \frac{217}{5}$$$

So $$\vec{d} = \left(-\frac{93}{5}, 31, \frac{217}{5}\right)$$. Factor out $$\frac{1}{5}$$:

$$$\vec{d} = \frac{1}{5}(-93, 155, 217)$$$

Notice that $$-93 = -3 \times 31$$, $$155 = 5 \times 31$$, and $$217 = 7 \times 31$$, so:

$$$\vec{d} = \frac{31}{5}(-3, 5, 7)$$$

The direction vector can be simplified to $$\vec{d} = (-3, 5, 7)$$ by scaling.

The acute angle $$\theta$$ between line $$L$$ and the $$y$$-axis (direction vector $$\vec{j} = (0, 1, 0)$$) is given by:

$$$\cos \theta = \frac{|\vec{d} \cdot \vec{j}|}{|\vec{d}| \cdot |\vec{j}|}$$$

Since $$|\vec{j}| = 1$$:

$$$\vec{d} \cdot \vec{j} = (-3)(0) + (5)(1) + (7)(0) = 5$$$

$$$|\vec{d}| = \sqrt{(-3)^2 + 5^2 + 7^2} = \sqrt{9 + 25 + 49} = \sqrt{83}$$$

Thus:

$$$\cos \theta = \frac{|5|}{\sqrt{83}} = \frac{5}{\sqrt{83}}$$$

Then:

$$$\cos^2 \theta = \left(\frac{5}{\sqrt{83}}\right)^2 = \frac{25}{83}$$$

Now compute $$415 \cos^2 \theta$$:

$$$415 \times \frac{25}{83} = \frac{415 \times 25}{83}$$$

Since $$415 \div 83 = 5$$:

$$$\frac{415}{83} \times 25 = 5 \times 25 = 125$$$

Therefore, $$415 \cos^2 \theta = 125$$.