The converse of $$((\sim p) \wedge q) \Rightarrow r$$ is

If four distinct points with position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ are coplanar, then $$[\vec{a} \,\,\vec{b} \,\,\vec{c}]$$ is equal to :

Let $$\mathrm{A}=\{1,3,4,6,9\}$$ and $$\mathrm{B}=\{2,4,5,8,10\}$$. Let $$\mathrm{R}$$ be a relation defined on $$\mathrm{A} \times \mathrm{B}$$ such that $$\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.$$ and $$\left.b_{1} \leq a_{2}\right\}$$. Then the number of elements in the set R is :

Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :