If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$,then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and the projection of $\vec{c}$ on $\vec{a}$ is $1$. Then,the projection of $\vec{c}$ on $\vec{b}$ equals:

The angle between two diagonals of a cube is:

Let $a, b, c \in \mathbb{R}$ be such that $a^{2} + b^{2} + c^{2} = 1$. If $a \cos \theta = b \cos \left(\theta + \frac{2\pi}{3}\right) = c \cos \left(\theta + \frac{4\pi}{3}\right)$ where $\theta = \frac{\pi}{9}$,then the angle between the vectors $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$ is:

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-

If the vectors $\bar{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\bar{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\bar{c}=p \hat{i}+\hat{j}+q \hat{k}$ are mutually orthogonal,then $(p, q)$ is equal to