If $\mathbf{a}=(x+2 y-3) \hat{\mathbf{i}}+(2 x-y+3) \hat{\mathbf{j}}$ and $\mathbf{b}=(3 x-2 y) \hat{\mathbf{i}} +(x-y+1) \hat{\mathbf{j}}$ are two vectors such that $\mathbf{a}=2 \mathbf{b}$, then $y-5 x=$

$7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}},-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are the position vectors of the points $A, B, C$ and $D$ respectively. If $p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $A B C D$, then $p+q+r=$

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 $p=$