Vector Equation of a Line

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.