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 $\mathbf{M})=$
- 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 $2 x-z=$
The point which lies on the plane passing through the point $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ is
If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, then the component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is
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