If $\bar{a}$ and $\bar{b}$ are mutually perpendicular unit vectors,then $(3\bar{a}+2\bar{b}) \cdot (5\bar{a}-6\bar{b}) = $

If $a = 4 \hat{i} + 6 \hat{j}$,$b = 3 \hat{j} + 4 \hat{k}$,and $c$ is the projection vector of $a$ on $b$,then $c$ and $|c|$ respectively are

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

The value of $c$ such that for all real $x$, the vectors $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$ make an obtuse angle is:

Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be four vectors such that $\vec{a}$ is perpendicular only to $\vec{c}$. If the vector $\vec{b}$ is parallel to $(\vec{c}-\vec{d})$,then $\vec{c}$ is equal to:

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$,$\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$,$\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then $x^2+y^2=$