If $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

The points $A(5, -1, 1)$,$B(7, -4, 7)$,$C(1, -6, 10)$,and $D(-1, -3, 4)$ are vertices of a:

If $\overrightarrow{a} \cdot \overrightarrow{b} = -|\overrightarrow{a}||\overrightarrow{b}|$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is (in $^{\circ}$)

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

The constant value $(\lambda + \mu)$ for which the lines $\vec{r} = (2\hat{i} + \hat{j} + \hat{k}) + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = (\hat{i} + \hat{j} - 3\hat{k}) + \mu(\hat{j} + 2\hat{k})$ intersect each other is equal to (where $\lambda$ and $\mu$ are parameters).

The value of $\theta \in (0, \pi/2)$ for which vectors $\vec{a} = (\sin \theta)\hat{i} + (\cos \theta)\hat{j}$ and $\vec{b} = \hat{i} - \sqrt{3}\hat{j} + 2\hat{k}$ are perpendicular is