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 :

Let the position vectors of three vertices of a triangle be $4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}$ and $2 \vec{p}-\vec{q}+2 \vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to :

Let $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :