Let the position vectors of points $A, B$ and $C$ in $∆ABC$ be $\widehat{i}+\widehat{j}+2\widehat{k}, \widehat{i}+2\widehat{j}+\widehat{k}$ and $2\widehat{i}+\widehat{j}+\widehat{k}$, respectively. Let $l_1, l_2$ and $l_3$ be the lengths of the perpendiculars drawn from the orthocentre $O$ on the sides $AB, BC$ and $CA$, respectively. Then $\left(l_1+l_2+l_3\right)$ is equal to

If $D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$, respectively, of a triangle $ABC$ and $\lambda$ is a scalar, such that $\overrightarrow{AD}+\frac{2}{3}\overrightarrow{BE}+\frac{1}{3}\overrightarrow{CF}=\lambda \overrightarrow{AC}$, then $\lambda$ is equal to

The position vectors of the points $A, B$, and $C$ are $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$ and $2\widehat{i}+3\widehat{j}+2\widehat{k}$, respectively. If $A$ is chosen as the origin, then the position vectors of $B$ and $C$ are