Area of ABC on a plane 
In the above question could anyone please explain me what they have done.
 A: I am sorry but I do not understand the suggested solution. 
Anyway, this is an alternative approach.
The cartesian equation of the plane $E$ is 
$$2x-y-2z+5=0$$
Therefore the distance of $A$ from that plane is
$$|AB|=\frac{|2(-4)-(2)-2(2)+5|}{\sqrt{2^2+(-1)^2+(-2)^2}}=3.$$
Moreover
$$|AC|=\|(-4-(-3),2-0,2-4)\|=\sqrt{1^2+2^2+(-2)^2}=3.$$
Now note that $AB \perp AC$, hence the required area is
$$\frac{|AB|\cdot|AC|}{2}=\frac{9}{2}.$$
A: Observe that
$$F:\;\;(-4,2,2)+\lambda(1,2,0)+\mu(1,0,1)$$
As $\;E,\,F\;$ are parallel planes, their distance $\;\mathcal D\;$ is the distance between any point in $\;E\;$ to the plane $\;F\;$, so:
$$\mathcal D^2=\min_{\lambda,\mu}d^2\left((-1,1,1)\,,\,\,(\lambda+\mu-4,\,2\lambda+2,\,\mu+2)\right)=$$
$$=\min\left[(\lambda+\mu-3)^2+(2\lambda+1)^2+(\mu+1)^2\right]=9\;\text{(why?)}$$
and thus the minimal distance is $\;3\;$
Since $\;||AB||\;$ is the above minimal distance and $\;C\;$ is on $\;F\;$ , the triangle $\;\Delta ABC\;$ is a straight angle one, with $\;\angle CAB=90^\circ\;$, and thus this triangle's area is simply
$$\frac{AC\cdot AB}2=\frac{3\cdot3}2=\frac92$$
