$\text{A straight line drawn from the point}P(1,3,2)\text{parallel to the line}\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}\text{ intersects the}$

$\text{plane}{\mathrm{ L}}_1:x-y+3z=6\text{ at the point}\mathrm{ Q}.\text{Another straight line which passes through}Q\text{and is perpendicular to the}$

$\text{ plane}{\mathrm{ L}}_1\text{intersects the plane}{\mathrm{ L}}_2:2x-y+z=-4\text{ at the point}\mathrm{ R}.\text{Then which of the following statements}$

$\text{is(are) TRUE?}$

The correct statements are:

1. The length of the line segment $PQ$ is $\sqrt{6}$

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 the point $P(1,3,2)$. A 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 are $⟨1,2,1⟩$.

Therefore, the equation of the line passing through $P(1,3,2)$ and parallel to the given line is:

$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1}=t$

Any arbitrary point on this line can be written in parametric form as:

$Q=(1+t,3+2t,2+t)$

Since point $Q$ lies on the plane $L_1:x-y+3z=6$, we substitute the coordinates of $Q$ into the equation of plane $L_1$:

$(1+t)-(3+2t)+3(2+t)=6$

$1+t-3-2t+6+3t=6$

$2t+4=6\Rightarrow 2t=2\Rightarrow t=1$

Substituting $t=1$ back to find coordinates of $Q$:

$Q=(1+1,3+2(1),2+1)=(2,5,3)$

Step 2: Calculate the length of segment $PQ$

Using the 3D distance formula between $P(1,3,2)$ and $Q(2,5,3)$:

$PQ=\sqrt{(2-1{)}^2+(5-3{)}^2+(3-2{)}^2}$

$PQ=\sqrt{1^2+2^2+1^2}=\sqrt{1+4+1}=\sqrt{6}$

Hence, the statement "The length of the line segment $PQ$ is $\sqrt{6}$" is TRUE.

Step 3: Find the coordinates of point $R$

A line passes through $Q(2,5,3)$ and is perpendicular to plane $L_1:x-y+3z=6$.

The normal vector to plane $L_1$ is $\overrightarrow{n}=⟨1,-1,3⟩$. Thus, the equation of this perpendicular line is:

$\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}=k$

Any point on this line can be expressed as $R=(2+k,5-k,3+3k)$.

Since point $R$ lies on plane $L_2:2x-y+z=-4$, we substitute $R$ into plane $L_2$:

$2(2+k)-(5-k)+(3+3k)=-4$

$4+2k-5+k+3+3k=-4$

$6k+2=-4\Rightarrow 6k=-6\Rightarrow k=-1$

Thus, the coordinates of $R$ are:

$R=(2-1,5-(-1),3+3(-1\left)\right)=(1,6,0)$

Step 4: Find the centroid of triangle $PQR$

The vertices of triangle $PQR$ are $P(1,3,2)$, $Q(2,5,3)$, and $R(1,6,0)$.

The formula for the centroid of a triangle in 3D space is:

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$

Calculating each coordinate:

$x_{\text{centroid}}=\frac{1+2+1}{3}=\frac{4}{3}$

$y_{\text{centroid}}=\frac{3+5+6}{3}=\frac{14}{3}$

$z_{\text{centroid}}=\frac{2+3+0}{3}=\frac{5}{3}$

Therefore, the centroid is $(\frac{4}{3},\frac{14}{3},\frac{5}{3})$.

Hence, the statement "The centroid of the triangle $PQR$ is $(\frac{4}{3},\frac{14}{3},\frac{5}{3})$" is TRUE.