$a, b, c$ are non-coplanar vectors. If $a+3 b+4 c=x(a-2 b+3 c)+y(a+5 b-2 c)+z(6 a+14 b+4 c)$,then $x+y+z=$

If $\bar{a} = (x + 2y - 3) \bar{i} + (2x - y + 3) \bar{j}$ and $\bar{b} = (3x - 2y) \bar{i} + (x - y + 1) \bar{j}$ are two vectors such that $\bar{a} = 2 \bar{b}$, then $y - 5x =$

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}$ be two vectors such that $\vec{a} \cdot \vec{b}=1$, $\cos(\theta) = \frac{1}{3}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, and the components of $\vec{b}$ with respect to $(\hat{i}, \hat{j}, \hat{k})$ are integers. Then the number of possible vectors that represent $\vec{b}$ is

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

If $|\vec{c}|^2 = 60$ and $\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,then a value of $\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ is

The position vector of a point $C$ with respect to $B$ is $i + j$ and that of $B$ with respect to $A$ is $i - j$. The position vector of $C$ with respect to $A$ is