If four distinct points with position vectors → a , → b , → c and →...

Given that $\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c}, \overset{\to}{d}$ are coplanar.

$⇒ \overset{\to}{b} - \overset{\to}{a}, \overset{\to}{c} - \overset{\to}{a}, \overset{\to}{d} - \overset{\to}{a}$ are also coplanar.

⇒ \left[\begin{cases}\overset{\to}{b} - \overset{\to}{a} & \overset{\to}{c} - \overset{\to}{a} & \overset{\to}{d} - \overset{\to}{a}\end{cases}\right] = 0

$⇒ \left(\overset{\to}{b} - \overset{\to}{a}\right) . \left(\left(\overset{\to}{c} - \overset{\to}{a}\right) \times \left(\overset{\to}{d} - \overset{\to}{a}\right)\right) = 0$

$⇒ \left(\overset{\to}{b} - \overset{\to}{a}\right) . \left(\overset{\to}{c} \times \overset{\to}{d} - \overset{\to}{c} \times \overset{\to}{a} - \overset{\to}{a} \times \overset{\to}{d}\right) = 0 ( Since\overset{\to}{a} \times \overset{\to}{a} = 0 )$

$⇒ \overset{\to}{b} . \left(\overset{\to}{c} \times \overset{\to}{d}\right) - \overset{\to}{b} . \left(\overset{\to}{c} \times \overset{\to}{a}\right) - \overset{\to}{b} . \left(\overset{\to}{a} \times \overset{\to}{d}\right) - \overset{\to}{a} . \left(\overset{\to}{c} \times \overset{\to}{d}\right) + \overset{\to}{a} . \left(\overset{\to}{c} \times \overset{\to}{a}\right) + \overset{\to}{a} . \left(\overset{\to}{a} \times \overset{\to}{d}\right) = 0$

⇒ \left[\begin{cases}\overset{\to}{b} & \overset{\to}{c} & \overset{\to}{d}\end{cases}\right] - \left[\begin{cases}\overset{\to}{b} & \overset{\to}{c} & \overset{\to}{a}\end{cases}\right] - \left[\begin{cases}\overset{\to}{b} & \overset{\to}{a} & \overset{\to}{d}\end{cases}\right] - \left[\begin{cases}\overset{\to}{a} & \overset{\to}{c} & \overset{\to}{d}\end{cases}\right] = 0

We know that \left[$\begin{cases}\overset{\to}{x} & \overset{\to}{y} & \overset{\to}{z}\end{cases}$\right] = \left[$\begin{cases}\overset{\to}{y} & \overset{\to}{z} & \overset{\to}{x}\end{cases}$\right] = \left[$\begin{cases}\overset{\to}{z} & \overset{\to}{x} & \overset{\to}{y}\end{cases}$\right] and \left[$\begin{cases}\overset{\to}{x} & \overset{\to}{y} & \overset{\to}{z}\end{cases}$\right] = - \left[$\begin{cases}\overset{\to}{x} & \overset{\to}{z} & \overset{\to}{y}\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}\overset{\to}{b} & \overset{\to}{c} & \overset{\to}{d}\end{cases}$\right] + \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{d}\end{cases}$\right] + \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{d} & \overset{\to}{c}\end{cases}$\right] = \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}\overset{\to}{c} & \overset{\to}{d} & \overset{\to}{b}\end{cases}$\right] + \left[$\begin{cases}\overset{\to}{b} & \overset{\to}{d} & \overset{\to}{a}\end{cases}$\right] + \left[$\begin{cases}\overset{\to}{d} & \overset{\to}{c} & \overset{\to}{a}\end{cases}$\right] = \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}$\right]</p><p>Hence, the required answer is \left[\begin{cases}\overset{\to}{d} & \overset{\to}{c} & \overset{\to}{a}\end{cases}\right] + \left[\begin{cases}\overset{\to}{b} & \overset{\to}{d} & \overset{\to}{a}\end{cases}\right] + \left[\begin{cases}\overset{\to}{c} & \overset{\to}{d} & \overset{\to}{b}\end{cases}\right]