Consider the vectors $\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}$, $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-5 \hat{i}-2 \hat{j}+3 \hat{k}$. If $l, m$ and $n$ are the lengths of the projections of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$ and $\vec{c}$ on $\vec{a}$ respectively, then:

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$

Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$, then $|\vec{a}|^2=$

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 $\vec a = \hat i - \hat j,$ $\vec b = \hat i + \hat j + \hat k$ and $\vec c$ be a vector such that $\vec a \times \vec c + \vec b = 0$ and $\vec a \cdot \vec c = 4$,then ${\left| {\vec c} \right|^2}$ is equal to

$AB=a$ 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