$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

If $\overrightarrow{a} = \hat{i} + 2\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$,such that $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{c}$ is equal to:

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are orthogonal to each other, then the locus of the point $(x, y)$ is

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ where $|\bar{a}|=4, |\bar{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$,then $|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$ has the value

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is