If the vectors $\mathbf{B C}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{C D}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ represent two adjacent sides of a parallelogram ABCD and $\theta$ is the angle between its diagonals $\mathbf{A C}$ and $\mathbf{B D}$, then $\tan \theta=$

A plane $\pi$ passing through the point $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is parallel to the plane which passes through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and perpendicular to the vector $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$. Then, the cartesian equation of $\pi$ is

Let $\mathbf{a}=\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ and $\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ be three vectors for some integer $\lambda$. If the volume of the parallelopiped with $\mathbf{a}, \mathbf{b}, \mathbf{c}$ as coterminous edges is 61 cubic units, then the number of possible values of $\lambda$ is

If the variance of the data $2,3,5,8,12$ is $\sigma^2$ and the mean deviation from the median for this data is $M$, then $\sigma^2-M=$