Let $$L_1$$ be the line of intersection of the planes given by the equations

$$2x + 3y + z = 4 \quad \text{and} \quad x + 2y + z = 5.$$

Let $$L_2$$ be the line passing through the point $$P(2, -1, 3)$$ and parallel to $$L_1$$. Let $$M$$ denote the plane given by the equation

$$2x + y - 2z = 6.$$

Suppose that the line $$L_2$$ 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?

The equations of the two given planes are
  $$\pi_1: \; 2x + 3y + z = 4,$$
  $$\pi_2: \; x + 2y + z = 5.$$

1. Direction vector of their line of intersection $$L_1$$
For two intersecting planes the direction vector is the cross-product of their normals.
  Normal of $$\pi_1 : \; \mathbf{n}_1 = (2,3,1)$$
  Normal of $$\pi_2 : \; \mathbf{n}_2 = (1,2,1)$$
$$\mathbf{d}= \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & 1\\ 1 & 2 & 1 \end{vmatrix} = (1,\,-1,\,1).$$

2. One point on $$L_1$$
Put $$z=0$$ in both plane equations:
$$2x + 3y = 4, \qquad x + 2y = 5.$$ Solving, $$y=6,\; x=-7.$$
So $$A(-7,\,6,\,0)$$ lies on $$L_1.$$ Hence $$L_1:\; (x,y,z)=(-7,6,0)+t(1,-1,1).$$

3. Equation of $$L_2$$
$$L_2$$ passes through $$P(2,-1,3)$$ and is parallel to $$L_1,$$ so it has the same direction vector $$\mathbf{d}=(1,-1,1).$$
$$L_2:\; (x,y,z)=(2,-1,3)+t(1,-1,1).$$

4. Point $$Q$$ where $$L_2$$ meets plane $$M: 2x + y - 2z = 6$$
Substitute $$x=2+t,\; y=-1-t,\; z=3+t$$ into the plane:
$$2(2+t)+(-1-t)-2(3+t)=6$$ $$4+2t-1-t-6-2t=6$$ $$-3 - t = 6 \;\Longrightarrow\; t=-9.$$ Therefore $$Q:(x,y,z)=\bigl(2-9,\,-1+9,\,3-9\bigr)=(-7,\,8,\,-6).$$

5. Length of segment $$PQ$$
$$\overrightarrow{PQ}=Q-P=(-9,\,9,\,-9).$$
$$|PQ|=\sqrt{(-9)^2+9^2+(-9)^2}=9\sqrt{3}.$$ Option A is TRUE.

6. Foot $$R$$ of the perpendicular from $$P$$ to plane $$M$$
Plane $$M: 2x + y - 2z - 6 = 0$$ has normal $$\mathbf{n}=(2,1,-2).$$
For a point $$P(x_0,y_0,z_0)$$, the foot is $$R = P - \frac{ax_0 + by_0 + cz_0 + d}{a^2+b^2+c^2}\,(a,b,c).$$ Here $$ax_0+by_0+cz_0+d = 2(2)+1(-1)+(-2)(3)-6=-9,$$ $$a^2+b^2+c^2=4+1+4=9,$$ so the factor is $$\dfrac{-9}{9}=-1.$$ Hence $$R = (2,-1,3) - (-1)(2,1,-2) = (2+2,\,-1+1,\,3+(-2)) = (4,\,0,\,1).$$

7. Length of segment $$QR$$
$$\overrightarrow{QR}=R-Q=(11,\,-8,\,7).$$
$$|QR|=\sqrt{11^2+(-8)^2+7^2}=\sqrt{234}\;(\approx 15.329).$$ Option B states the length is 15, so it is FALSE.

8. Area of $$\triangle PQR$$
Using vectors $$\overrightarrow{PQ}=(-9,9,-9)$$ and $$\overrightarrow{PR}=R-P=(2,1,-2),$$
$$\overrightarrow{PQ}\times\overrightarrow{PR} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -9&9&-9\\ 2&1&-2 \end{vmatrix} =(-9,\,-36,\,-27).$$ Magnitude $$= \sqrt{(-9)^2+(-36)^2+(-27)^2} = \sqrt{2106}=3\sqrt{234}.$$ Area $$\triangle PQR=\frac12\bigl| \overrightarrow{PQ}\times\overrightarrow{PR}\bigr| =\frac12\,(3\sqrt{234})=\frac{3}{2}\sqrt{234}.$$ Option C is TRUE.

9. Acute angle between $$PQ$$ and $$PR$$
$$\overrightarrow{PQ}\cdot\overrightarrow{PR}=(-9)(2)+9(1)+(-9)(-2)=9.$$ $$|PQ|=9\sqrt3,\quad |PR|=\sqrt{2^2+1^2+(-2)^2}=3.$$ $$\cos\theta = \frac{9}{9\sqrt3\cdot3}= \frac{1}{3\sqrt3}.$$ Option D gives $$\cos^{-1}\!\left(\dfrac{1}{2\sqrt3}\right),$$ hence it is FALSE.

Conclusion
Correct statements: Option A and Option C.