Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \cdot \mathbf{a}=0$, $\mathbf{r} \cdot \mathbf{b}=-2$ and $\mathbf{r} \cdot \mathbf{c}=6$, then $\mathbf{r} \cdot(\beta \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=$

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{d}$ is a vector perpendicular to both $\mathbf{a}, \mathbf{b}$ and $|\mathbf{d} \times \mathbf{c}|=14$, then $|\mathbf{d} \cdot \mathbf{c}|=$

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 $p+q+r=$