If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector,then $(a \cdot b)b + (a \cdot c)c + \frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = $

Let $\bar{a} = 2\hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} + 2\hat{j} - \hat{k}$,and vector $\bar{c}$ be coplanar with $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{a}$,then $\bar{c}$ is:

If $|a| = 3$ and $|b| = 4$,then a value of $\lambda$ for which $a + \lambda b$ is perpendicular to $a - \lambda b$ is

If the angle between the vectors $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$ is obtuse,then $\lambda \in$

If $A(3,2,3)$, $B(1,4,6)$, and $C(7,4,5)$ are the three vertices of a parallelogram $ABCD$, then the angle between its diagonal through $D$ and the side $DC$ is

$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, -\bar{i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+\bar{k}$ are the position vectors of the points $A, B, C, D$ respectively. If $p \bar{i}+q \bar{j}+r \bar{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $ABCD$, then $p+q+r=$