Orthocentre and Area of Triangle Formed by Line Intersections in Three Dimensions

To determine which of the given statements are correct, we will solve the problem in a step-by-step manner.

Step 1: Finding the Coordinates of Point $S$

The line $L$ passes through the points $P(1, 2, -1)$ and $Q(2, 3, 1)$. The direction vector of line $L$ is given by: $\vec{d} = \vec{PQ} = (2-1)\hat{i} + (3-2)\hat{j} + (1-(-1))\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$

Any general point $S$ on the line $L$ can be expressed in parametric form using parameter $\lambda$: $S = (1 + \lambda, 2 + \lambda, -1 + 2\lambda)$

The point $R$ is given as $R(4, -1, 5)$. The vector $\vec{RS}$ is: $\vec{RS} = S - R = (1 + \lambda - 4)\hat{i} + (2 + \lambda - (-1))\hat{j} + (-1 + 2\lambda - 5)\hat{k} = (\lambda - 3)\hat{i} + (\lambda + 3)\hat{j} + (2\lambda - 6)\hat{k}$

Since $S$ is the foot of the perpendicular from $R$ to the line $L$, we have $\vec{RS} \perp \vec{d}$, which means their dot product must be zero: $\vec{RS} \cdot \vec{d} = 0$ $(\lambda - 3)(1) + (\lambda + 3)(1) + (2\lambda - 6)(2) = 0$ $\lambda - 3 + \lambda + 3 + 4\lambda - 12 = 0$ $6\lambda - 12 = 0 \implies \lambda = 2$

Substituting $\lambda = 2$ into the coordinates of $S$: $S = (1 + 2, 2 + 2, -1 + 2(2)) = (3, 4, 3)$

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Step 2: Finding the Coordinates of Point $T$

We are given that point $S$ divides the segment $PT$ internally in the ratio $|PS| : |ST| = 1 : 2$. Using the section formula for internal division: $S = \frac{2P + 1T}{1 + 2} \implies 3S = 2P + T \implies T = 3S - 2P$

Substituting the coordinates of $S(3, 4, 3)$ and $P(1, 2, -1)$: $T = 3(3, 4, 3) - 2(1, 2, -1) = (9 - 2, 12 - 4, 9 + 2) = (7, 8, 11)$

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Step 3: Calculating the Area of Triangle $PRT$

The base of $\triangle PRT$ lies along the segment $PT$, and its height is the perpendicular distance $|RS|$.

1. Calculate length of segment $PS$: $\vec{PS} = S - P = (3 - 1)\hat{i} + (4 - 2)\hat{j} + (3 - (-1))\hat{k} = 2\hat{i} + 2\hat{j} + 4\hat{k}$ $|PS| = \sqrt{2^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$

2. Calculate total base length $|PT|$: Since $|PS| : |ST| = 1 : 2$, the total length $|PT| = |PS| + |ST| = 3|PS|$: $|PT| = 3(2\sqrt{6}) = 6\sqrt{6}$

3. Calculate perpendicular height $|RS|$: $\vec{RS} = S - R = (3 - 4)\hat{i} + (4 - (-1))\hat{j} + (3 - 5)\hat{k} = -\hat{i} + 5\hat{j} - 2\hat{k}$ $|RS| = \sqrt{(-1)^2 + 5^2 + (-2)^2} = \sqrt{1 + 25 + 4} = \sqrt{30}$

4. Calculate Area of $\triangle PRT$: $\text{Area}(\triangle PRT) = \frac{1}{2} \times |PT| \times |RS| = \frac{1}{2} \times 6\sqrt{6} \times \sqrt{30}$ $\text{Area}(\triangle PRT) = 3 \times \sqrt{180} = 3 \times 6\sqrt{5} = 18\sqrt{5}$

Thus, Statement (D) is correct and Statement (C) is false.

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Step 4: Finding the Orthocentre of Triangle $PRT$

Let $H$ be the orthocentre of $\triangle PRT$.

Since $RS \perp PT$, $RS$ is the altitude from vertex $R$ to the base $PT$. Therefore, the orthocentre $H$ lies on the line passing through $R$ and $S$.

The line containing $RS$ can be parametrized using vector $\vec{SR} = (4 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - 3)\hat{k} = \hat{i} - 5\hat{j} + 2\hat{k}$: $H = (3 + m, 4 - 5m, 3 + 2m)$

Now, the altitude from vertex $P$ to side $RT$ must be perpendicular to $RT$.

• Vector $\vec{RT} = T - R = (7 - 4)\hat{i} + (8 - (-1))\hat{j} + (11 - 5)\hat{k} = 3\hat{i} + 9\hat{j} + 6\hat{k}$, which is parallel to $\hat{i} + 3\hat{j} + 2\hat{k}$.
• Vector $\vec{PH} = H - P = (3 + m - 1)\hat{i} + (4 - 5m - 2)\hat{j} + (3 + 2m - (-1))\hat{k} = (2 + m)\hat{i} + (2 - 5m)\hat{j} + (4 + 2m)\hat{k}$.

Since $PH \perp RT$: $\vec{PH} \cdot (\hat{i} + 3\hat{j} + 2\hat{k}) = 0$ $(2 + m)(1) + (2 - 5m)(3) + (4 + 2m)(2) = 0$ $2 + m + 6 - 15m + 8 + 4m = 0$ $16 - 10m = 0 \implies m = \frac{16}{10} = \frac{8}{5}$

Substituting $m = \frac{8}{5}$ into the expression for $H$: $H = \left(3 + \frac{8}{5}, 4 - 5\left(\frac{8}{5}\right), 3 + 2\left(\frac{8}{5}\right)\right) = \left(\frac{23}{5}, -4, \frac{31}{5}\right)$

Thus, Statement (A) is correct and Statement (B) is false.

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Conclusion:

The correct statements are A and D.