If $2 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}$ and $-\mathbf{i}+2 \mathbf{j}+\mathbf{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is

Let the vectors $\mathbf{A B}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{A C}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ be two sides of a $\triangle A B C$. If $G$ is the centroid of $\triangle A B C$, then $\frac{27}{7}|\mathbf{A G}|^2+5=$

If $(\alpha, \beta, \gamma)$ is a triad of real numbers satisfying $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\alpha(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\beta(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\gamma(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$, then $\alpha^2-\beta^2+\gamma^2=$

If $\theta$ is the angle between the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $a \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+b \hat{\mathbf{k}}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$