A straight line drawn from the point P(1,3, 2), parallel to the line $\frac{x-<!--\; -\; -->2}{1}=\frac{y-<!--\; -\; -->4}{2}=\frac{z-<!--\; -\; -->6}{1}$, intersects the plane $L_1:x-<!--\; -\; -->y+3z=6$ at the point Q. Another straight line which passes through Q and is perpendicular to the plane L₁ intersects the plane $L_2:2x-<!--\; -\; -->y+z=-<!--\; -\; -->4$ at the point R. Then which of the following statements is (are) TRUE?

The correct statements are:
1. The length of the line segment PQ is √6
2. The centroid of the triangle PQR is $(\frac{4}{3},\frac{14}{3},\frac{5}{3})$

Step 1: Find the coordinates of point Q
We are given a point P(1, 3, 2).
A straight line is drawn from P parallel to the line:
$\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$
The direction ratios of this line, and therefore any line parallel to it, are (1, 2, 1).
Thus, the equation of the line passing through P(1, 3, 2) in parametric form is:
$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1}=\lambda$
Any general point on this line can be represented as:
$(x,y,z)=(\lambda +1,2\lambda +3,\lambda +2)$
Since Q is the intersection of this line with the plane L₁: x - y + 3z = 6, we substitute the coordinates of the general point into the plane equation:
$(\lambda +1)-(2\lambda +3)+3(\lambda +2)=6$
Simplifying the equation:
$\lambda +1-2\lambda -3+3\lambda +6=6$
$2\lambda +4=6$
$2\lambda =2\Rightarrow \lambda =1$
Substituting λ = 1 back into the parametric coordinates, we get the coordinates of Q:
$Q=(1+1,2(1)+3,1+2)=(2,5,3)$

Step 2: Find the length of the line segment PQ
We have P(1, 3, 2) and Q(2, 5, 3). Using the distance formula:
$PQ=\sqrt{{(2-1)}^2+{(5-3)}^2+{(3-2)}^2}$
$PQ=\sqrt{1^2+2^2+1^2}=\sqrt{1+4+1}=\sqrt{6}$
Thus, the length of the line segment PQ is indeed √6. This statement is TRUE.

Step 3: Find the coordinates of point R
We are given that another line passes through Q(2, 5, 3) and is perpendicular to the plane L₁: x - y + 3z = 6.
The normal vector to L₁ is:
$\overrightarrow{n}=(1,-1,3)$
Since the line is perpendicular to the plane L₁, its direction is parallel to the normal vector.
Thus, the parametric equation of the line passing through Q(2, 5, 3) with direction ratios (1, -1, 3) is:
$\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}=\mu$
Any general point on this line can be written as:
$(x,y,z)=(\mu +2,-\mu +5,3\mu +3)$
This line intersects the plane L₂: 2x - y + z = -4 at point R. We substitute the general point into the equation of L₂:
$2(\mu +2)-(-\mu +5)+(3\mu +3)=-4$
Simplifying:
$2\mu +4+\mu -5+3\mu +3=-4$
$6\mu +2=-4$
$6\mu =-6\Rightarrow \mu =-1$
Substituting μ = -1 back to find R:
$R=(-1+2,-(-1)+5,3(-1)+3)=(1,6,0)$
Thus, the coordinates of R are (1, 6, 0). (Therefore, the statement that coordinates of R are (1, 6, 3) is FALSE.)

Step 4: Find the centroid of the triangle PQR
We have:
P = (1, 3, 2)
Q = (2, 5, 3)
R = (1, 6, 0)
The centroid G(x_c, y_c, z_c) of triangle PQR is given by:
$x_c=\frac{1+2+1}{3}=\frac{4}{3}$
$y_c=\frac{3+5+6}{3}=\frac{14}{3}$
$z_c=\frac{2+3+0}{3}=\frac{5}{3}$
Thus, the centroid of triangle PQR is:
$(\frac{4}{3},\frac{14}{3},\frac{5}{3})$
This statement is TRUE.