Find the value of $\lambda$ such that the vectors $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ are orthogonal.

The value of $\lambda$ for which the vectors $3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are parallel, is

The vectors from origin to the points $A$ and $B$ are $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ respectively, then the area of $\triangle O A B$ is equal to

For any vector $\overrightarrow{\mathbf{a}}$, the value of $(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}})^2+(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}})^2+(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}})^2$ is

If $|\overrightarrow{\mathbf{a}}|=10,|\overrightarrow{\mathbf{b}}|=2$ and $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=12$, then the value of $|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$ is