If three points $A$,$B$,and $C$ have position vectors $(1, x, 3)$,$(3, 4, 7)$,and $(y, -2, -5)$ respectively and if they are collinear,then $(x, y)$ is

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

The position vector of a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

Let $b = 3j + 4k$,$a = i + j$ and let $b_1$ and $b_2$ be component vectors of $b$ parallel and perpendicular to $a$ respectively. If $b_1 = \frac{3}{2}i + \frac{3}{2}j$,then $b_2 = $

If $\vec{a}$ and $\vec{b}$ are two collinear vectors,then which of the following is incorrect?

If $A(2 \hat{i} + \hat{j} - \hat{k})$, $B(\lambda \hat{i} + 5 \hat{j} + 4 \hat{k})$, $C(-4 \hat{i} + 3 \hat{j} + 2 \hat{k})$ and $D(-\hat{i} - 2 \hat{j} + 3 \hat{k})$ are four points in space such that $\overrightarrow{AB} = x \overrightarrow{AC} + y \overrightarrow{AD}$ for some real numbers $x \neq 0, y \neq 0$, then $17(\lambda + 9) =$ ?