If $$\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w $$   are non-coplanar vectors and $$p,\,q$$  are real numbers, then the equality $$\left[ {3\overrightarrow u \,p\overrightarrow v \,p\overrightarrow w } \right] - \left[ {p\overrightarrow v \,p\overrightarrow w \,q\overrightarrow u } \right] - \left[ {2\overrightarrow w \,q\overrightarrow v \,q\overrightarrow u } \right] = 0$$           holds for :

Solution :
$$\because \,\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w $$   are non-coplanar vectors
$$\eqalign{ & \therefore \left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] \ne 0 \cr & {\text{Now, }}\left[ {3\overrightarrow u ,\,p\overrightarrow v ,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v ,\,p\overrightarrow w ,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow w ,\,q\overrightarrow v ,\,q\overrightarrow u } \right] = 0 \cr & \Rightarrow 3{p^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] - pq\left[ {\overrightarrow v ,\,\overrightarrow w ,\,\overrightarrow u } \right] - 2{q^2}\left[ {\overrightarrow w ,\,\overrightarrow v ,\,\overrightarrow u } \right] = 0 \cr & \Rightarrow 3{p^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] - pq\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] + 2{q^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] = 0 \cr & \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] = 0 \cr & \Rightarrow 3{p^2} - pq + 2{q^2} = 0 \cr & \Rightarrow 2{p^2} + {p^2} - pq + \frac{{{q^2}}}{4} + \frac{{7{q^2}}}{4} = 0 \cr & \Rightarrow 2{p^2} + {\left( {p - \frac{q}{2}} \right)^2} + \frac{7}{4}{q^2} = 0 \cr & \Rightarrow p = 0,\,q = 0,\,p = \frac{q}{2} \cr} $$
This is possible only when $$p = 0,\,q = 0$$
$$\therefore $$  There is exactly one value of $$\left( {p,\,q} \right)$$