Let L be the straight line joining the points P(1, 2, -1) and Q(2,...

The direction ratios of line $L$ passing through $P(1, 2, -1)$ and Q(2, 3, 1)$are$(2-1, 3-2, 1-(-1)) = (1, 1, 2)$.<br><br>The equation of line$L$is$\dfrac{x-1}{1} = \dfrac{y-2}{1} = \dfrac{z+1}{2} = \lambda$.<br><br>Any point on$L$can be written as$S(\lambda+1, \lambda+2, 2\lambda-1)$.<br><br>Since$S$is the foot of the perpendicular from$R(4, -1, 5)$to$L$, the direction ratios of$RS$are$(\lambda-3, \lambda+3, 2\lambda-6)$.<br><br>As$RS \perp L$, the dot product of their direction ratios is zero:<br><br>$1(\lambda-3) + 1(\lambda+3) + 2(2\lambda-6) = 0$<br><br>$6\lambda - 12 = 0 \Rightarrow \lambda = 2$<br><br>Thus, the coordinates of$S$are$(3, 4, 3)$.<br><br>The length of$PS$is$\sqrt{(3-1)^2 + (4-2)^2 + (3-(-1))^2} = \sqrt{4+4+16} = 2\sqrt{6}$.<br><br>The length of the altitude$RS$is$\sqrt{(3-4)^2 + (4-(-1))^2 + (3-5)^2} = \sqrt{1+25+4} = \sqrt{30}$.<br><br>Since$S$divides$PT$internally in the ratio$1:2$, we have:<br><br>$S = \dfrac{T + 2P}{3} \Rightarrow T = 3S - 2P$<br><br>$T = 3(3, 4, 3) - 2(1, 2, -1) = (7, 8, 11)$<br><br>The length of the base$PT$is$3|PS| = 6\sqrt{6}$.<br><br>The area of$\triangle PRT$is$\dfrac{1}{2} \times |PT| \times |RS| = \dfrac{1}{2} \times 6\sqrt{6} \times \sqrt{30} = 18\sqrt{5}$.<br><br>The orthocentre$H$lies on the altitude$RS$. The line$RS$passes through$S(3, 4, 3)$and has direction ratios proportional to$R-S = (1, -5, 2)$.<br><br>Let the coordinates of$H$be$(3+t, 4-5t, 3+2t)$.<br><br>Since$H$is the orthocentre,$PH \perp RT$.<br><br>The direction ratios of$RT$are$(7-4, 8-(-1), 11-5) = (3, 9, 6)$, which is proportional to$(1, 3, 2)$.<br><br>The direction ratios of$PH$are$(3+t-1, 4-5t-2, 3+2t-(-1)) = (t+2, 2-5t, 2t+4)$.<br><br>Taking the dot product of$PH$and$RT$:<br><br>$1(t+2) + 3(2-5t) + 2(2t+4) = 0$<br><br>$t + 2 + 6 - 15t + 4t + 8 = 0$<br><br>$16 - 10t = 0 \Rightarrow t = \dfrac{8}{5}$<br><br>Substituting$t = \dfrac{8}{5}$into the coordinates of$H$, we get:<br><br>$H = \left(3 + \dfrac{8}{5}, 4 - 5\left(\dfrac{8}{5}\right), 3 + 2\left(\dfrac{8}{5}\right)\right) = \left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$<br><br>Answer: The orthocentre of the triangle$PRT$is \left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$; The area of the triangle $PRT$ is $18\sqrt{5}$