Find the position vector of the point which divides the join of the points with position vectors $2\vec{a} - 3\vec{b}$ and $3\vec{a} - 2\vec{b}$ externally in the ratio $2:3$.

If the vectors $\vec{AB} = -3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$, then the length of the median through $A$ is

Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$ be three vectors such that $|\vec{a}-\vec{b}+\vec{c}|=\sqrt{141}$. If $y_1$ and $y_2$ are the values of $y$ satisfying the given condition,then $|y_1-y_2|=$

If $m_1, m_2, m_3$ and $m_4$ are respectively the magnitudes of the vectors $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$, then the correct order of $m_1, m_2, m_3$ and $m_4$ is

The ratio in which $\hat{i}+2 \hat{j}+3 \hat{k}$ divides the join of $-2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $7 \hat{i}-\hat{k}$ is