Area of Triangle in 3D

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.

The direction vector of the line is $$\vec d = (2, -1, 2)$$, and its magnitude is:

$$|\vec d| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$$

Given that the base $$BC = 6$$ and the $$\text{Area of } \triangle ABC = 3\sqrt{10}$$, the perpendicular height $$h$$ from point $$A$$ to the line is:

$$h = \frac{2 \times \text{Area}}{BC} = \frac{2 \times 3\sqrt{10}}{6} = \sqrt{10}$$

Let a fixed point on the line be $$P(\alpha+1, 1, 3)$$.

Given $$A(\alpha, 2, \alpha)$$, the vector $$\overrightarrow{PA}$$ is:

$$\overrightarrow{PA} = (\alpha - (\alpha+1))\hat i + (2 - 1)\hat j + (\alpha - 3)\hat k = -\hat i + \hat j + (\alpha - 3)\hat k$$

Now, find the cross product $$\overrightarrow{PA} \times \vec d$$:

$$\overrightarrow{PA} \times \vec d = \begin{vmatrix}\hat i & \hat j & \hat k \\ -1 & 1 & \alpha-3 \\ 2 & -1 & 2 \end{vmatrix}$$  $$= \big(2 - (-1)(\alpha-3)\big)\hat i - \big(-2 - 2(\alpha-3)\big)\hat j + \big(1 - 2\big)\hat k$$  $$= (\alpha - 1)\hat i - (4 - 2\alpha)\hat j - \hat k$$

The perpendicular distance $$h$$ is also given by the formula $$h = \frac{|\overrightarrow{PA} \times \vec d|}{|\vec d|}$$. Therefore:

$$\frac{\sqrt{(\alpha - 1)^2 + (4 - 2\alpha)^2 + (-1)^2}}{3} = \sqrt{10}$$

Squaring both sides:

$$(\alpha - 1)^2 + (4 - 2\alpha)^2 + 1 = 90$$

Expand the terms:

$$(\alpha^2 - 2\alpha + 1) + (16 - 16\alpha + 4\alpha^2) + 1 = 90$$  $$5\alpha^2 - 18\alpha + 18 = 90$$ $$5\alpha^2 - 18\alpha - 72 = 0$$

Factor the quadratic equation:

$$5\alpha^2 - 30\alpha + 12\alpha - 72 = 0$$ $$5\alpha(\alpha - 6) + 12(\alpha - 6) = 0$$ $$(5\alpha + 12)(\alpha - 6) = 0$$

Since $$\alpha \in \mathbb{Z}$$ (it must be an integer), we select:

$$\alpha = 6$$

$$\alpha^2 = 6^2 = 36$$