A) Exactly two values of (p, q)
B) More than two but not all values of (p, q)
C) All values of (p, q)
D) Exactly one value of (p, q)
Correct Answer: D
Solution :
[d] \[\because \vec{u},\,\vec{v},\,\vec{w}\] are non-coplanar vectors \[\because \left[ \vec{u},\,\vec{v},\,\vec{w} \right]\ne 0\] Now, \[\left[ 3\vec{u},\,p\vec{v},\,p\vec{w} \right]-\left[ p\vec{v},p\vec{w},q\vec{u} \right]-\left[ 2\vec{w},q\vec{v},q\vec{u} \right]=0\] \[\Rightarrow 3{{p}^{2}}\left[ \vec{u},\vec{v},\vec{w} \right]-pq\left[ \vec{v},\vec{w},\vec{u} \right]-2{{q}^{2}}\left[ \vec{w},\vec{v},\vec{u} \right]=0\] \[\Rightarrow 3{{p}^{2}}\left[ \vec{u},\vec{v},\vec{w} \right]-pq\left[ \vec{u},\vec{v},\vec{w} \right]-2{{q}^{2}}\left[ \vec{u},\vec{v},\vec{w} \right]\] \[\Rightarrow (3{{p}^{2}}-pq+2{{q}^{2}})\left[ \vec{u},\vec{v},\vec{w} \right]=0\] \[\Rightarrow 3{{p}^{2}}-pq+2{{q}^{2}}=0\] \[\Rightarrow 2{{p}^{2}}+{{p}^{2}}-pq+\frac{{{q}^{2}}}{4}+\frac{7{{q}^{2}}}{4}=0\] \[\Rightarrow 2{{p}^{2}}+{{\left( p-\frac{q}{2} \right)}^{2}}+\frac{7}{4}{{q}^{2}}=0\Rightarrow p=0,q=0,\] \[p=q/2\] This is possible only when \[p=0,q=0\] \[\therefore \]There is exactly one value of (p, q).
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