The resultant of the two vectors having magnitude $2$ and $3$ is $1$. What is their cross product

For three vectors $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k})$, $\vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})$, if $\overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}})=0$, them value of $\mathrm{x}$ is. . . . . .. 

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} . \vec{B}=[\vec{A} \times \vec{B}]$. Then the value of $[\vec{A}-\vec{B}]$. will be :

A particle moves in the $x-y$ plane under the action of a force $\overrightarrow F $ such that the value of its linear momentum $(\overrightarrow P )$ at anytime t is ${P_x} = 2\cos t,\,{p_y} = 2\sin t.$ The angle $\theta $between $\overrightarrow F $ and $\overrightarrow P $ at a given time $t$. will be $\theta =$ ........... $^o$

Two forces ${\vec F_1} = 5\hat i + 10\hat j - 20\hat k$ and ${\vec F_2} = 10\hat i - 5\hat j - 15\hat k$ act on a single point. The angle between ${\vec F_1}$ and ${\vec F_2}$ is nearly ....... $^o$