The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

Let $O$ be the origin,and $\overline{OX}, \overline{OY}, \overline{OZ}$ be three unit vectors in the directions of the sides $QR, RP, PQ$,respectively,of a triangle $PQR$.
$(1)$ Find $|\overline{OX} \times \overline{OY}|$.
$[A] \sin(P+Q)$
$[B] \sin 2R$
$[C] \sin(P+R)$
$[D] \sin(Q+R)$
$(2)$ If the triangle $PQR$ varies,then find the minimum value of $\cos(P+Q) + \cos(Q+R) + \cos(R+P)$.
$[A] -\frac{5}{3}$
$[B] -\frac{3}{2}$
$[C] \frac{3}{2}$
$[D] \frac{5}{3}$
Select the correct options for $(1)$ and $(2)$.

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors. Suppose $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$. Then $\vec{a}$ is

If $a + b + c = 0,$ then which relation is correct?

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is:

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.