The line $$PQ$$ has direction vector
$$\vec d=(1,1,2)$$
Hence,
$$L:\ (x,y,z)=(1,2,-1)+\lambda(1,1,2)$$
Let
$$S=(1+\lambda,2+\lambda,-1+2\lambda)$$
Since $$RS\perp L$$,
$$[(4,-1,5)-S]\cdot(1,1,2)=0$$
Therefore,
$$(3-\lambda)+(-3-\lambda)+2(6-2\lambda)=0$$
$$12-6\lambda=0$$
$$\lambda=2$$
Hence,
$$S=(3,4,3)$$
Now,
$$PS=\sqrt{(2)^2+(2)^2+(4)^2}=2\sqrt6$$
Given,
$$PS:ST=1:2$$
Therefore,
$$ST=4\sqrt6$$
and since $$P,S,T$$ are collinear,
$$S$$ divides $$PT$$ internally in the ratio $$1:2$$.
Using section formula,
$$S=\frac{2P+T}{3}$$
Hence,
$$T=3S-2P$$
$$=(9,12,9)-(2,4,-2)$$
$$=(7,8,11)$$
Now,
$$\overrightarrow{PT}=(6,6,12)=6(1,1,2)$$
and
$$\overrightarrow{RS}=(-1,5,-2)$$
Since
$$\overrightarrow{PT}\cdot\overrightarrow{RS}=0$$
the altitude from $$R$$ passes through $$S$$.
Also,
$$PR^2=(3)^2+(-3)^2+(6)^2=54$$
$$RT^2=(3)^2+(9)^2+(6)^2=126$$
$$PT^2=(6)^2+(6)^2+(12)^2=216$$
Since
$$54+126=180\ne216$$
the triangle is not right-angled.
The area is
$$\frac12|\overrightarrow{PR}\times\overrightarrow{PT}|$$
$$=\frac12|(3,-3,6)\times(6,6,12)|$$
$$=\frac12|(-72,0,36)|$$
$$=\frac12\sqrt{72^2+36^2}$$
$$=\frac12\sqrt{6480}$$
$$=18\sqrt5$$
Hence Statement 4 is FALSE.
To find the orthocentre, the altitude through $$R$$ is
$$R+\mu(-1,5,-2)$$
The altitude through $$P$$ is perpendicular to $$RT$$.
A point common to the two altitudes is
$$\left(\frac{23}{5},-4,\frac{31}{5}\right)$$
Hence Statement 1 is TRUE.
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