If $\mathbf{a}=\hat{\mathbf{i}}+p \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{b}=p \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are three vectors such that $|\mathbf{a} \times \mathbf{b}|=\mid \mathbf{a} \times \mathbf{c}$, then $p=$

If $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{d}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are four vectors, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=$

The variance of ungrouped data $2,12,3,11,5,10,6,7$, is

If $A$ and $B$ are events of a random experiment such that $P(A \cup B)=\frac{3}{4}, P(A \cap B)=\frac{1}{4}, P(\overline{\mathrm{~A}})=\frac{2}{3}$, then $P(\overline{\mathrm{~A}} \cap \mathrm{~B})=$