Three points whose position vectors are $a + b$,$a - b$,and $a + kb$ will be collinear if the value of $k$ is:

Let $\vec{a}, \vec{b}, \vec{c}$ be co-initial vectors and $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=3 \hat{i}+7 \hat{j}-\hat{k}$. Let $(\vec{a}, \vec{b})=\theta$ be an acute angle and $\vec{c}$ be the vector along the bisector of the angle $\theta$. If $\lambda, x, y \in R$,then $\vec{c}=$

Suppose that $\bar{p}, \bar{q}$ and $\bar{r}$ are three non-coplanar vectors in $\mathbb{R}^3$. Let the components of a vector $\bar{s}$ along $\bar{p}, \bar{q}$ and $\bar{r}$ be $4, 3$ and $5$ respectively. If the components of this vector $\bar{s}$ along $(-\bar{p}+\bar{q}+\bar{r}), (\bar{p}-\bar{q}+\bar{r})$ and $(-\bar{p}-\bar{q}+\bar{r})$ are $x, y$ and $z$ respectively,then the value of $2x+y+z$ is

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:3$ internally and $Q$ is the mid-point of $AC$. If $\vec{AB} + \vec{AD} + \vec{BC} - 2\vec{DC} = \lambda \vec{PQ}$, then the value of $\lambda$ is

Find the unit vector in the direction of the sum of the vectors,$\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$.

Let the position vectors of the vertices $A, B$ and $C$ of a triangle be $2 \hat{i}+2 \hat{j}+\hat{k}$,$\hat{i}+2 \hat{j}+2 \hat{k}$ and $2 \hat{i}+\hat{j}+2 \hat{k}$ respectively. Let $l_1, l_2$ and $l_3$ be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides $AB, BC$ and $CA$ respectively,then $l_1^2+l_2^2+l_3^2$ equals: