If $$\alpha, \beta$$ and $$\gamma$$ are the angles between the vectors $$\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$$, and $$\overrightarrow{\mathrm{R}}$$ and $$\alpha=90^{\circ}$$ as shown in figure. the product of $$(\vec{Q} \times \vec{R}) \cdot \vec{Q}$$ is equal to

The sides of a parallelogram are represented by vectors $$\vec{p}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$$ and $$\vec{q}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$$. Then, the area of the parallelogram is

The resultant of two forces acting at an angle of 120$$^\circ$$ is 10 kg-W and is perpendicular to one of the forces. That force is