Match List - I with List - II.
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... \infty = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... \infty = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + ... \infty = \beta, $ then $ \frac{\alpha}{\beta} $ is equal to:
Let $\alpha$ be a solution of $x^2 + x + 1 = 0$, and for some $a$ and $b$ in $\mathbb{R}$, $ \begin{bmatrix} 1 & 16 & 13 \\-1 & -1 & 2 \\-2 & -14 & -8 \end{bmatrix} \begin{bmatrix} 4 \\a \\b \end{bmatrix} = \begin{bmatrix} 0 \\0 \\0 \end{bmatrix}. $ If $\frac{4}{\alpha^4} + \frac{m} {\alpha^a} + \frac{n}{\alpha^b} = 3$, then $m + n$ is equal to _____.
Let $ A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\4 & 6 + 2p & 8 + 3p + 2q \\6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} $ If $ \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n, \, m, n \in \mathbb{N}, $ then $ m + n $ is equal to: