If $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are the position vectors of the vertices $A, B, C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is

Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors satisfying $|\mathbf{a}-\mathbf{b}|^2+|\mathbf{a}-\mathbf{c}|^2=10$. Then,

Statement (I): $|\mathbf{a}+2 \mathbf{b}|^2+|2 \mathbf{a}+\mathbf{c}|^2=2$

Statement (II) : $|2 a+3 b|^2+|3 a+2 c|^2=10$

Which of the above statements is (are) true?

If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ are the position vectors of the points $\mathbf{A}$ and $\mathbf{B}$ respectively, $\mathbf{C}$ divides $\mathbf{A B}$ in the ratio $2: 3$ and $\mathbf{M}$ is the mid-point of $A B$, then 5 (position vector of $\mathbf{C})-2($ position vector of $\mathbf{M})=$

30. If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are the non-coplanar vectors and $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c},-4 \mathbf{a}+5 \mathbf{b}-6 \mathbf{c}, x \mathbf{a}-9 \mathbf{b}+z \mathbf{c}$ are collinear points, then $2 x-z=$