Take any two pairs of the given points and make vectors out of them. For example, the vector from A to B is
[tex]\vec v_1 = \langle 1,4,5\rangle - \langle 1,1,1\rangle = \langle0,3,4\rangle[/tex]
and the vector from A to C is
[tex]\vec v_2 = \langle -3,-2,0\rangle - \langle1,1,1\rangle = \langle-4,-3,-1\rangle[/tex]
These vectors lie in the same plane (the one we want). We can get a third vector that is normal to the plane by taking their cross product (details omitted).
[tex]\vec n = \vec v_1 \times \vec v_2 = \langle 9,-16,12 \rangle[/tex]
If [tex]\vec u = \langle x,y,z\rangle[/tex] is an arbitrary vector, then the vector from any of the points A, B, or C to [tex]\vec u[/tex] will lie in our plane. That is, if we start from A,
[tex](\vec u - \langle1,1,1\rangle) \cdot \vec n = 0[/tex]
and this reduces to the equation of the plane,
[tex]\langle x - 1, y - 1, z - 1 \rangle \cdot \langle 9, -16, 12 \rangle = 0[/tex]
[tex]9 (x - 1) - 16 (y - 1) + 12 (z - 1) = 0[/tex]
[tex]\boxed{9x - 16y + 12z = 5}[/tex]
This follows immediately from the cross product identity
[tex]\|\vec x \times \vec y\| = \|\vec x\| \|\vec y\| \sin(\theta)[/tex]
where [tex]\theta[/tex] is the angle between [tex]\vec x[/tex] and [tex]\vec y[/tex]. The left side corresponds to the area of the parallelogram spanned by [tex]\vec x[/tex] and [tex]\vec y[/tex]; half of this area would be that of a triangle. (see attached)
In our case, we have
[tex]\|\vec n\| = \|\vec v_1 \times \vec v_2\| = \sqrt{481}[/tex]
so the area of ABC is [tex]\boxed{\dfrac{\sqrt{481}}2}[/tex].
We already know the first vector, [tex]v_1[/tex].
The vector from C to B is
[tex]\vec v_3 = \langle 1,4,5 \rangle - \langle -3,-2,0 \rangle = \langle 4, 6, 5 \rangle[/tex]
Recall the dot product identity,
[tex]\vec x \cdot \vec y = \|\vec x\| \|\vec y\| \cos(\theta)[/tex]
Then
[tex]\vec v_1 \cdot \vec v_3 = \|\vec v_1\| \|\vec v_3\| \cos(\theta) \implies \cos(\theta) = \dfrac{38}{5\sqrt{77}} \implies \theta \approx \boxed{29.9914^\circ}[/tex]
