If $\vec{u}, \vec{v}, \bar{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality…
If $\vec{u}, \vec{v}, \bar{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality $\left[\begin{array}{llll}3 \vec{u} & p \vec{v} & p \vec{w}\end{array}\right]-\left[\begin{array}{lll}p \vec{v} & \vec{w} & q \vec{u}\end{array}\right]-\left[\begin{array}{lll}2 \vec{w} & q \vec{v} & q \vec{u}\end{array}\right]=0$ holds for
exactly one value of $(p, q)$
exactly two values of $(p, q)$
more than two but not all values of $(p, q)$
all values of $(p, q)$
Solution
$
\begin{aligned}
& \left(3 p^2-p q+2 q^2\right)\left[\begin{array}{lll}
\vec{u} & \vec{v} & \bar{w}
\end{array}\right]=0 \\
& \text { But }\left[\begin{array}{lll}
\vec{u} & \vec{v} & \bar{w}
\end{array}\right] \neq 0 \\
& 3 p^2-p q+2 q^2=0 \\
& 2 p^2+p^2-p q+\left(\frac{q}{2}\right)^2+\frac{7 q^2}{4}=0 \Rightarrow 2 p^2+\left(p-\frac{q}{2}\right)^2+\frac{7}{4} q^2=0
\end{aligned}
$
$
\Rightarrow p=0, q=0, p=\frac{q}{2}
$
This possible only when $p=0, q=0$ exactly one value of $(p, q)$