$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$,with $|\lambda| \leq 5$,are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units,then $5-6 \lambda$ is equal to $.........$.

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

Let the lines $L_1: \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_2: \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$. The unit vector perpendicular to both $L_1$ and $L_2$ is:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ are two vectors,and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{c}$ is perpendicular to $\vec{b}$,then find the value of $\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k})$.

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

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 $(2)$.