Consider a $\triangle ABC$ where $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$,then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

Let $\overline{a}=3 \hat{i}-\alpha \hat{j}+\hat{k}$ and $\overline{b}=\hat{i}+\alpha \hat{j}+3 \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overline{a}$ and $\overline{b}$ is $8 \sqrt{3}$ sq. units,then $\overline{a} \cdot \overline{b}$ is equal to

If $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ and $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$,and $f(t) = a \cdot b$,then $f^{\prime}(1)$ is equal to

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If $p \times q = p \times r$ and $p \cdot q = p \cdot r$,then $\ldots . . .$.

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to: