Area of ABC on a plane

Question

In the above question could anyone please explain me what they have done.

Answer 1

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 $$\frac{|AB|\cdot|AC|}{2}=\frac{9}{2}.$$

Answer 2

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

$$\frac{AC\cdot AB}2=\frac{3\cdot3}2=\frac92$$