Let $\mathbf{a}$ be a vector in the plane containing vectors $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{a}$ is perpendicular to $\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and its projection on $\mathbf{b}$ is $3 \sqrt{6}$, then $|\mathbf{a}|^2=$

The cartesian equation of the plane passing through the point $(1,-2,3)$ and perpendicular to the vector $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$, is

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, $\mathbf{d}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be four vectors and let $l=\mathbf{b} \cdot \mathbf{c}$ and $m=\mathbf{c} \cdot \mathbf{a}$. Then, $[m \mathbf{b}+l \mathbf{a} \mathbf{b d}]=$

If $\bar{x}$ is the mean of $n$ observations $x_1, x_2, \ldots ., x_n$ then the mean of the absolute deviations of these observations from $\bar{x}$ is