If the angle between the vectors $a$ and $b$ is $\theta$ and $a \cdot b = \cos \theta$,then the true statement is:

Let $\bar{u}, \bar{v}, \bar{w}$ be vectors such that $|\bar{u}|=1, |\bar{v}|=2, |\bar{w}|=3$. If the projection of $\bar{v}$ on $\bar{u}$ is equal to that of $\bar{w}$ on $\bar{u}$,and the vectors $\bar{v}, \bar{w}$ are perpendicular to each other,then $|\bar{u}-\bar{v}+\bar{w}|=$

$A$ unit vector in the plane of the vectors $2i + j + k$ and $i - j + k$ and orthogonal to $5i + 2j + 6k$ is

If the position vectors of three points $A, B, C$ are $\hat{i}+2\hat{j}+\hat{k}$,$2\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ respectively,then the perpendicular distance of the point $C$ from the line $AB$ is

If $|\vec{a}|=1, |\vec{b}|=2, |\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$,then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is:

$AB=a$ 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