Let $ L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2} $ and $ L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1} $ be two lines. Let $ L_3 $ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $ L_1 $ and $ L_2 $. If $ L_3 $ intersects $ L_1 $, then $ 5 \alpha = 11 \beta - 8 \gamma $ equals :

Let $ P $ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $ P $ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $ P $ is :

Let $ A = \begin{bmatrix} a_{ij} \end{bmatrix} = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix} $. If $ A_{ij} $ is the cofactor of $ a_{ij} $, $ C_{ij} = \sum\limits_{k=1}^{2} a_k A_{jk} , 1 \leq i, j \leq 2 $, and $ C=[C_{ij}] $, then $ 8|C| $ is equal to :

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11^th term is :