Let $x_1, x_2, ..., x_{10}$ be ten observations such that $\sum\limits_{i=1}^{10} (x_i - 2) = 30$, $\sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98$, $\beta > 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta$, then $\frac{\beta\mu}{\sigma^2}$ is equal to :

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+7 \hat{j}+3 \hat{k}$. Let $\mathrm{L}_1 : \overrightarrow{\mathrm{r}}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \vec{a}, \mathrm{\lambda} \in \mathbf{R}$ and $\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{j}+\hat{k})+\mu \vec{b}, \mu \in \mathrm{R}$ be two lines. If the line $\mathrm{L}_3$ passes through the point of intersection of $\mathrm{L}_1$ and $L_y$ and is parallel to $\vec{a}+\vec{b}$, then $L_3$ passes through the point :

Let the area of the region

$ (x, y) : 2y \leq x^2 + 3,\ y + |x| \leq 3, \ y \geq |x - 1| $ be $ A $. Then $ 6A $ is equal to :

Let $ \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}, \ \vec{b} = 3\hat{i} - 5\hat{j} + \hat{k} $ and $ \vec{c} $ be a vector such that $ \vec{a} \times \vec{c} = \vec{a} \times \vec{b} = \vec{c} \times \vec{b} $ and $ (\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168 $. Then the maximum value of $|\vec{c}|^2$ is :