If $\mathbf{a}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}, \mathbf{b}=-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ are three vectors such that $(\mathbf{b} \times \mathbf{c}) \times \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then $x+y-z=$
The following data represents the frequency distribution of 20 observations
$$ \begin{array}{ccccccc} \hline x_i & 3 & 4 & 5 & 8 & 10 & 11 \\ \hline f_i & \alpha+2 & (\alpha-1)^2 & 4 & \alpha-1 & 2 & \alpha \\ \hline \end{array} $$
Then, its mean deviation about the mean is
If $l, m$ represent any two elements (identical or different) of the set $\{1,2,3,4,5,6,7\}$, then the probability that $l x^2+m x+1>0 \forall x \in R$ is
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