If vectors satisfy the condition $|a - c| = |b - c|$,then $(b - a) \cdot \left( c - \frac{a + b}{2} \right)$ is equal to

In a triangle $ABC,$ right-angled at the vertex $A,$ if the position vectors of $A, B,$ and $C$ are respectively $3\hat{i} + \hat{j} - \hat{k},$ $-\hat{i} + 3\hat{j} + p\hat{k},$ and $5\hat{i} + q\hat{j} - 4\hat{k},$ then the point $(p, q)$ lies on a line

Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$, $\vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-4 \hat{j}-12 \hat{k}$ be three vectors. If $\vec{p}$ is the projection of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection of $\vec{c}$ on $\vec{a}$, then $13 \vec{p}=$ (in $\vec{q}$)

If $a$ and $b$ are two vectors,then $(a \times b)^2$ equals

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is