Find the vector joining the points $P(2, 3, 0)$ and $Q(-1, -2, -4)$ directed from $P$ to $Q$.

Let $u, v$ and $w$ be three vectors in $R^3$. Then,any vector $z \in R^3$ can be written as $z = au + bv + cw$ for some scalars $a, b$ and $c$ if and only if:

If $M_1, M_2, M_3$ and $M_4$ are respectively the magnitudes of the vectors $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -3\hat{i} - 4\hat{j} - 4\hat{k}$,$\vec{a}_3 = -\hat{i} + \hat{j} - \hat{k}$,and $\vec{a}_4 = -\hat{i} + 3\hat{j} + \hat{k}$,then the correct order of $M_1, M_2, M_3$ and $M_4$ is:

If $\vec{a}=(2x+y)\hat{i}+3\hat{j}+9\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-(x-y)\hat{k}$ are two collinear vectors,then $x^3+27y^3=$

If $G$ is the centroid of the $\triangle ABC$,then $\vec{GA} + \vec{GB} + \vec{GC}$ is equal to

Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \vec{a}+2 \vec{b}$ and $3 \vec{a}+\lambda \vec{b}$ are collinear, then the set of all possible values of $\lambda$ is