Three Dimensional Geometry

Concept:Find foot of perpendicular from $R$ to line $L$ (through $P$ and $Q$), then locate $T$ on $L$ such that $S$ divides $PT$ internally in $1:2$. Compute area of $\triangle PRT$ and verify orthocenter using altitudes.Explanation:Line $L$: $\frac{x-1}{1}=\frac{y-2}{1}=\frac{z+1}{2}=\lambda$.General point: $(\lambda+1,\lambda+2,2\lambda-1)$.Let $S$ be foot from $R(4,-1,5)$ to $L$.Vector $\overrightarrow{RS}$: $(\lambda-3,\lambda+3,2\lambda-6)$.Direction of $L$: $(1,1,2)$.Perpendicular condition: dot product $=0$:$(\lambda-3)(1)+(\lambda+3)(1)+(2\lambda-6)(2)=0$$\Rightarrow 6\lambda-12=0 \Rightarrow \lambda=2$.Thus $S(3,4,3)$.$S$ divides $PT$ internally in $1:2$. Let $T(a,b,c)$.Section formula: $\frac{1\cdot a + 2\cdot 1}{3}=3$, etc.$\Rightarrow a=7,\; b=8,\; c=11$. So $T(7,8,11)$.Area of $\triangle PRT$: $RS \perp PT$ (since $S$ is foot from $R$ to $L$ and $PT$ lies on $L$).$PT=\sqrt{(6)^2+(6)^2+(12)^2}=6\sqrt{6}$.$RS=\sqrt{(-1)^2+5^2+(-2)^2}=\sqrt{30}$.Area $=\frac12 \cdot PT\cdot RS = \frac12 \cdot 6\sqrt{6}\cdot \sqrt{30}=18\sqrt{5}$.Hence option (D) is correct.For orthocenter $H$ of $\triangle PRT$, since $RS$ is one altitude, $H$ lies on $RS$.Equation of $RS$: $\frac{x-4}{1}=\frac{y+1}{-5}=\frac{z-5}{2}$.Check option (A): $H\left(\frac{23}{5},-4,\frac{31}{5}\right)$ lies on $RS$?Compute $\overrightarrow{PH}=\left(\frac{18}{5},-6,\frac{36}{5}\right)$ and $\overrightarrow{RT}=(-3,-9,-6)$.Dot product: $\left(\frac{18}{5}\right)(-3)+(-6)(-9)+\left(\frac{36}{5}\right)(-6)= -\frac{54}{5}+54-\frac{216}{5}=0$.Thus $PH \perp RT$, so $H$ lies on altitude from $P$. Hence $H$ is orthocenter. Option (A) is correct.Option (B) gives $S$, not orthocenter. Option (C) gives $6\sqrt{5}$, which is incorrect.Answer:Options (A) and (D) are TRUE.