Vector Algebra and Calculus

The cross product of two vectors is zero if they are parallel or anti-parallel. A is along the x-axis (i). Vector B must be a multiple of i. -4i is parallel to i.

$\left( \vec { A } \times \vec { B }  \right) =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 5 & 7 & -3 \ 15 & 21 & a \end{matrix} \right| =0\quad or\quad 5a+45=0\quad or\quad or\quad 7a+63=0\quad \Rightarrow a=-9$

By definition, C = A x B is perpendicular to both A and B. It is not necessarily perpendicular to the sum or difference of A and B. However, C is definitely NOT perpendicular to itself (C x C = 0, but C dot C = |C|^2, which is non-zero). Option C is the incorrect statement.

(A + B) dot (A - B) = |A|^2 - |B|^2. A = 4i + 5j - 6k, |A|^2 = 16 + 25 + 36 = 77. B = 2i - 3j + 4k, |B|^2 = 4 + 9 + 16 = 29. 77 - 29 = 48.

For vectors to be parallel, their components must be proportional: 2/6 = 3/alpha = 4/beta. 1/3 = 3/alpha => alpha = 9. 1/3 = 4/beta => beta = 12.

(B) $\overrightarrow{A} \times \overrightarrow{B}= \begin{vmatrix} \widehat{i}& \widehat{j} &\widehat{k}\\  4& 6  & 0\\ 2 & 3 & 0\end{vmatrix} = \widehat{i}(0-0)-\widehat{j}(0-0)+\widehat{k}(12-12)=0 $

$  \overrightarrow{A} \times \overrightarrow{B}=0   \overrightarrow{A}  \ and\  \overrightarrow{B} $ are parallel.
$  \overrightarrow{B} \times \overrightarrow{C}=0  \Rightarrow  \overrightarrow{B}  \ and \  \overrightarrow{C} $ are parallel 
 $\Rightarrow \vec{A} \parallel \vec{C}$

The vector perpendicular to the $i$ & $j$ plane would be along the unit vector $k$.
Another vector in $i$ & $j$ plane can be perpendicular to $\vec{F}$
$(3\widehat{i} +4\widehat{j}) \perp (4\widehat{i} -3\widehat{j}) $
But,  from the options only $7\widehat{k} \perp (4\widehat{i} -3\widehat{j})$

A vector $\displaystyle \vec{A}$ is parallel to an another vector $\displaystyle \vec{B}$ if it can be written as
$\displaystyle \vec{A}= m\vec{B}$ where $m$ is a constant.
Here, $\displaystyle \vec{A}=\left ( \hat{i}-\hat{j}+2\hat{k} \right )=\frac{1}{3}\left ( 3\hat{i}-3\hat{j}+6\hat{k} \right )$
or $\displaystyle \vec{A}=\frac{1}{3}\vec{B}$
This implies that $\vec A || \displaystyle \vec{B}$ and magnitude of $\displaystyle \vec{A}$ is $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

$\vec{a}\times\vec{b}=\begin{pmatrix}x _1\hat {i}+y _1\hat {j}\end{pmatrix}\times\begin{pmatrix}x _2\hat {i}+y _2\hat {j}\end{pmatrix}$

$\;\;\;\;\;\;\;\;\;\;\;=x _1y _2\hat {k}-x _2y _1\hat {k}=\vec{0}\Rightarrow x _1y _2=x _2y _1$

The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B} = \hat{n} |A| |B| \sin x $, where  $\hat{n}$ is the unit vector perpendicular to both A and B vectors and x is the angle between them.
Here in this question the vector product of $\vec{P1}$ and $\vec{P2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2) the $\sin$ of the angle between them is zero.
From the given options, the vector parallel to the given vector  $\hat{i}$ is  $-4\hat{i}$. 

$\displaystyle \displaystyle \overrightarrow A .  \overrightarrow B = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow B$
and, $\overrightarrow A .  \overrightarrow C = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow C$
Also, $\overrightarrow B \times \overrightarrow C $ is perpendicular to both $\vec { B } \ and  \ \vec { C }$
$Thus, \overrightarrow { A } ||\, (\overrightarrow { B } \times \overrightarrow { C } )$

Magnitude of a vector A can be given as $ \sqrt{\overrightarrow{A}.\overrightarrow{A}} $. 
Hence, statement A is wrong .
Also, if $ \overrightarrow{a} = m \overrightarrow{b}$, taking dot product with b vector on both sides,
$ \overrightarrow{a}.\overrightarrow{b} = m \  b^2 \Rightarrow m = \dfrac {\overrightarrow{a}.\overrightarrow{b}}{b^2} $
Hence , statement B is correct.

$\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$  and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$  is given.