Intersection and Distance Analysis of Line and Plane with Triangle Geometry

To determine which statements are true, let us analyze the problem step-by-step:

Step 1: Find the direction vector of the line $L_1$ and equation of line $L_2$

The line $L_1$ is the line of intersection of the planes: $2x + 3y + z = 4 \quad \text{and} \quad x + 2y + z = 5$

The normal vectors to these planes are $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + 2\hat{j} + \hat{k}$, respectively.

The direction vector $\vec{d}$ of the line $L_1$ is parallel to $\vec{n}_1 \times \vec{n}_2$: $\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(2-1) + \hat{k}(4-3) = \hat{i} - \hat{j} + \hat{k}$

Since $L_2$ passes through the point $P(2, -1, 3)$ and is parallel to $L_1$, its parametric equation is: $\vec{r} = (2 + \lambda)\hat{i} + (-1 - \lambda)\hat{j} + (3 + \lambda)\hat{k}$

Step 2: Find the coordinates of point $Q$ and length $PQ$

Point $Q$ is the intersection of $L_2$ with the plane $M: 2x + y - 2z = 6$. Substituting the parametric coordinates into the plane equation: $2(2 + \lambda) + (-1 - \lambda) - 2(3 + \lambda) = 6$ $4 + 2\lambda - 1 - \lambda - 6 - 2\lambda = 6$ $-3 - \lambda = 6 \implies \lambda = -9$

Thus, the length of the line segment $PQ$ is: $PQ = \sqrt{(\lambda)^2 + (-\lambda)^2 + (\lambda)^2} = |\lambda|\sqrt{3} = 9\sqrt{3}$ Therefore, Option (A) is TRUE.

Step 3: Find the distance $PR$ and the length of $QR$

$R$ is the foot of the perpendicular drawn from $P(2, -1, 3)$ to the plane $M$. The perpendicular distance $PR$ from point $P$ to the plane $M$ is: $PR = \frac{|2(2) + 1(-1) - 2(3) - 6|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|4 - 1 - 6 - 6|}{\sqrt{9}} = \frac{|-9|}{3} = 3$

Since $PR \perp M$ and the segment $QR$ lies in the plane $M$, the triangle $\Delta PQR$ is a right-angled triangle with the right angle at $R$. By the Pythagorean theorem: $QR = \sqrt{PQ^2 - PR^2} = \sqrt{(9\sqrt{3})^2 - 3^2} = \sqrt{243 - 9} = \sqrt{234} = 3\sqrt{26} \neq 15$ Therefore, Option (B) is FALSE.

Step 4: Find the area of $\Delta PQR$

The area of the right-angled triangle $\Delta PQR$ is: $\text{Area}(\Delta PQR) = \frac{1}{2} \times PR \times QR = \frac{1}{2} \times 3 \times \sqrt{234} = \frac{3}{2}\sqrt{234}$ Therefore, Option (C) is TRUE.

Step 5: Find the angle between $PQ$ and $PR$

Let $\theta$ be the acute angle between $PQ$ and $PR$. In the right-angled triangle $\Delta PQR$: $\cos\theta = \frac{PR}{PQ} = \frac{3}{9\sqrt{3}} = \frac{1}{3\sqrt{3}}$ $\theta = \cos^{-1}\left(\frac{1}{3\sqrt{3}}\right) \neq \cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$ Therefore, Option (D) is FALSE.

---

Conclusion:

The correct statements are (A) and (C).