Triangle area, planes, distance between parallel planes and perpendicular planes

Using a vector method, find the area of the triangle with vertices $P_1$, $P_2$ and $P_3$.

Give your answer in the form $\frac{a\sqrt{b}}{c}$, where $a, b, c \in \Z^+$ and $c = a + 1$.

The points $P_1$, $P_2$ and $P_3$ lie in the plane $\Pi_1$.

Find the Cartesian equation of this plane.

Use the following information to answer part c and part d.

Now consider the plane $\Pi_2$ with equation $2x - y + z = 6$.

Show that the plane $\Pi_3$ given by $\frac{d - 1.5z}{3} = x - \frac{1}{2}y$, where $d \in \R$, is parallel to the plane $\Pi_2$.

Determine the values of $d$ so that the minimum distance between plane $\Pi_2$ and plane $\Pi_3$ is $2\sqrt{6}$.

Consider the plane $\Pi_2$, and the plane $\Pi_4$ given by $p^2x - (p + 2)y - 5z = 1$ for $p \in \R$.

Find the possible values of $p$ for these two planes to be perpendicular.