Let $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. The vector $\mathbf{x}$ such that $\mathbf{x} \cdot \mathbf{c}=60$ and perpendicular to both $\mathbf{a}, \mathbf{b}$ is

The shortest distance between the line $\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})$ and the plane $\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ is

For the following frequency distribution, the variance is approximately equal to

$$ \begin{array}{cccccc} \hline \begin{array}{c} \text { Class } \\ \text { Interval } \end{array} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text { Frequency } & 4 & 1 & 10 & 3 & 2 \\ \hline \end{array} $$

If the mean of the discrete distribution $8,9,6,5, x, 4$, 6, 5 is 6 , then its standard deviation (nearest to two decimal places) is