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 $ \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 :

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :