If $\mathbf{a}=\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$ are two vectors, then the component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+p \hat{\mathbf{k}}$ be two vectors.

If $(\mathbf{a}, \mathbf{b})=60^{\circ}$, then $p=$

$A, B, C$ and $D$, are any four points. If $E$ and $F$ are mid-points of $A C$ and $B D$ respectively, then $\mathbf{A B}+\mathbf{C B}+\mathbf{C D}+\mathbf{A D}=$

The four points whose position vectors are given by $2 a+3 b-c, a-2 b+3 c, 3 a+4 b-2 c$ and $a-6 b+6 c$ are