Find the value of $6(M+m)$ Let $q$, $r$ are unit vectors such that $\vec p = \vec q \times \vec p + \vec r$.  $M$ is the maximum and $m$ is the minimum value of $[\vec p \vec q \vec r]$, then we have to find the value of $6(M+m)$.
Should $\vec p$ be zero?  As in $\vec p = \vec q \times \vec p + \vec r$, $p$ cannot be equal to its perpendicular.
 A: 
$$\mathbf{p}=\mathbf{q} \times \mathbf{p}+\mathbf{r} \tag{1}$$

$\mathbf{p} \cdot \mathbf{p}$ gives
\begin{align}
  p^2 &= \mathbf{p} \cdot \mathbf{r} \\
  p^2 &= p\cos \theta
\end{align}

$$p=\cos \theta \tag{2}$$

$\mathbf{p} \cdot \mathbf{q}$ gives

$$\mathbf{p} \cdot \mathbf{q}=\mathbf{q} \cdot \mathbf{r}=\cos \phi \tag{3}$$

$\mathbf{p} \cdot \mathbf{r}$ gives
\begin{align}
  \mathbf{p} \cdot \mathbf{r} &= 1-\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} \\
  \mathbf{p} \times \mathbf{q} \cdot \mathbf{r} &= 1-\mathbf{p} \cdot \mathbf{r} \\
  &= 1-p^2 \\
  &= 1-\cos^2 \theta
\end{align}

$$\mathbf{p} \times \mathbf{q} \cdot \mathbf{r} = \sin^2 \theta \tag{4}$$

$\mathbf{p} \times \mathbf{q}$ gives
\begin{align}
  \mathbf{p} \times \mathbf{q} &=
  (\mathbf{q} \times \mathbf{p}) \times \mathbf{q}+
  \mathbf{r} \times \mathbf{q} \\
  &= q^2 \mathbf{p}-(\mathbf{p} \cdot \mathbf{q}) \, \mathbf{q}+
  \mathbf{r} \times \mathbf{q} \\
  &= \mathbf{p}-(\mathbf{p} \cdot \mathbf{q}) \, \mathbf{q}+
  \mathbf{r} \times \mathbf{q} \\
  \mathbf{p} \times \mathbf{q} \cdot \mathbf{r}
   &= \mathbf{p} \cdot \mathbf{r}-
      (\mathbf{p} \cdot \mathbf{q})(\mathbf{q} \cdot \mathbf{r})
\end{align}

$$\mathbf{p} \times \mathbf{q} \cdot \mathbf{r}=
\cos^2 \theta-\cos^2 \phi \tag{5}$$

Equating $(4)$ and $(5)$,
\begin{align}
  \sin^2 \theta &= \cos^2 \theta-\cos^2 \phi \\
  1-\cos^2 \theta &= \cos^2 \theta-\cos^2 \phi \\
  \cos^2 \theta &= \frac{1+\cos^2 \phi}{2} \\
  \cos^2 \theta-\cos^2 \phi &= \frac{1-\cos^2 \phi}{2} \\
  \mathbf{p} \times \mathbf{q} \cdot \mathbf{r} &= \frac{\sin^2 \phi}{2} \\
  m &= 0 \\
  M &= \frac{1}{2}
\end{align}
