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 $|\mathbf{d} \cdot \mathbf{c}|=$

The mean deviation from the mean of the discrete data $2,3,5,7,11,13,17,19,22$ is

Out of the given 25 consecutive position integers, three integers are drawn. If the least integer among given 25 integers is an odd number, then the probability that the sum of the three integers drawn is an even number is