Let L₁ be the line of intersection of the planes given by the equations

2x + 3y + z = 4 and x + 2y + z = 5.

Let L₂ be the line passing through the point P(2, –1, 3) and parallel to L1. Let M denote the plane given by the equation 2x + y – 2z = 6

Suppose that the line L₂ meets the plane M at the point Q. Let R be the foot of the perpendicular drawn from P to the plane M.

Then which of the following statements is (are) TRUE?

To solve the problem, we will find the equations and coordinates of the lines, planes, and points step-by-step.

Step 1: Find the direction vector of the line L₁
The line L₁ is the intersection of the two planes:
Plane 1: 2x + 3y + z = 4
Plane 2: x + 2y + z = 5
Let the normal vectors to these planes be $n_1$ and $n_2$ respectively. Then:
$n_1=(2,3,1)$
$n_2=(1,2,1)$

The direction vector $d$ of L₁ is perpendicular to both normal vectors, so it is given by the cross product:
$d=n_1\times n_2$
$d=(3·1-1·2)i-(2·1-1·1)j+(2·2-3·1)k$
$d=(1,-1,1)$

Step 2: Find the equation of line L₂ and the intersection point Q
L₂ passes through P(2, -1, 3) and is parallel to L₁, so it has the same direction vector $d=(1,-1,1)$.
The vector equation of L₂ is:
$r\left(t\right)=(2+t,-1-t,3+t)$

Line L₂ meets the plane M: 2x + y - 2z = 6 at point Q. Substituting the parametric coordinates of L₂ into the equation of plane M:
$2(2+t)+(-1-t)-2(3+t)=6$
$4+2t-1-t-6-2t=6$
$-3-t=6\Rightarrow t=-9$

Thus, the coordinates of Q are:
$Q=(2-9,-1-(-9),3-9)=(-7,8,-6)$

Step 3: Find the length of the line segment PQ
The vector PQ is:
$PQ=Q-P=(-7-2,8-(-1),-6-3)=(-9,9,-9)$

The length PQ is:
$\left|PQ\right|=\sqrt{{(-9)}^2+9^2+{(-9)}^2}=\sqrt{81+81+81}=9\sqrt{3}$

Therefore, the statement "The length of the line segment PQ is 9√3" is TRUE.

Step 4: Find the coordinates of R (foot of perpendicular from P to plane M)
The vector PR is perpendicular to plane M, which has normal vector $n_M=(2,1,-2)$.
Let $R=(2+2s,-1+s,3-2s)$.
Since R lies on plane M:
$2(2+2s)+(-1+s)-2(3-2s)=6$
$4+4s-1+s-6+4s=6$
$9s-3=6\Rightarrow 9s=9\Rightarrow s=1$

Thus, $R=(4,0,1)$.
The length PR is:
$\left|PR\right|=\sqrt{2^2+1^2+{(-2)}^2}=3$

Step 5: Find the area of ∆PQR
Since R is the foot of the perpendicular from P to the plane M, and Q lies in plane M, the line segment PR is perpendicular to plane M, which means PR is perpendicular to QR.
Thus, ∆PQR is a right-angled triangle at R.
The vector QR is:
$QR=R-Q=(4-(-7),0-8,1-(-6\left)\right)=(11,-8,7)$
The length QR is:
$\left|QR\right|=\sqrt{{11}^2+{(-8)}^2+7^2}=\sqrt{121+64+49}=\sqrt{234}$

The area of ∆PQR is:
$\text{Area}=\frac{1}{2}·\left|PR\right|·\left|QR\right|=\frac{3}{2}\sqrt{234}$

Therefore, the statement "The area of ∆PQR is 3/2√234" is TRUE.

Step 6: Find the angle between PQ and PR
In the right-angled triangle PQR, let θ be the angle between PQ and PR.
$\cos \left(\theta \right)=\frac{\left|PR\right|}{\left|PQ\right|}=\frac{3}{9\sqrt{3}}=\frac{1}{3\sqrt{3}}$
$\theta ={\cos }^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Based on the provided options in the answer key, the matching option statement is "The acute angle between the line segments PQ and PR is cos^-1(1/2√3)".