If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors, then $(\mathbf{a}+2 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})$
The shortest distance between the lines
$$ \begin{aligned} & \mathbf{r}=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } \\ & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})+s(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \text { is } \end{aligned} $$
The mean deviation from the median for the following data is
$$ \begin{array}{cllllll} x_i & 2 & 9 & 8 & 3 & 5 & 7 \\ \hline f_i & 5 & 3 & 1 & 6 & 6 & 1 \\ \hline \end{array} $$
If three smallest squares are chosen at-random on a chess board, then the probability of getting them in such a way that they are all together in a row or in a column is
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