Let $$P$$ be the plane such that it contains the straight line $$\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{1}$$ and is perpendicular to the plane $$x+2y+3z=4$$. Let $$P_1$$ be the plane which passes through the point $$(4,2,2)$$ and is parallel to $$P$$.

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

The plane $$P$$ contains the line

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

and is perpendicular to the plane

$$x+2y+3z=4$$

Let the direction ratios of the line be $$\vec{d}=(2,3,1)$$ and the normal to the given plane be $$\vec{n}=(1,2,3)$$

Since the line lies in plane $$P$$ and plane $$P$$ is perpendicular to the plane $$x+2y+3z=4$$

the normal to $$P$$ must be perpendicular to both $$\vec{d}$$ and $$\vec{n}$$

Hence,

$$\vec{N}=\vec{d}\times\vec{n}$$

$$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix}$$

$$=7\hat{i}-5\hat{j}+\hat{k}$$

Therefore, the equation of plane $$P$$ is

$$7(x-1)-5(y-3)+(z+2)=0$$

$$7x-5y+z=-10$$

Hence Statement 1 is TRUE.

Now,

$$P: 7x-5y+z+10=0$$

and since $$P_1$$ passes through $$(4,2,2)$$ and is parallel to $$P$$,

$$7(4)-5(2)+2+D=0$$

$$20+D=0$$

$$D=-20$$

Therefore,

$$P_1: 7x-5y+z-20=0$$

Distance between the planes is

$$\frac{|10-(-20)|}{\sqrt{7^2+(-5)^2+1^2}}$$

$$=\frac{30}{\sqrt{75}}$$

$$=\frac{30}{5\sqrt{3}}$$

$$=2\sqrt{3}$$

Hence Statement 2 is FALSE.

Distance of plane $$P$$ from the origin is

$$\frac{|10|}{\sqrt{75}}$$

$$=\frac{2}{\sqrt{3}}$$

$$=\frac{2\sqrt{3}}{3}$$

Hence Statement 3 is FALSE.

For the plane

$$2x+2y+z=3$$

the normal vector is

$$\vec{N}_1=(2,2,1)$$

The acute angle between the planes equals the angle between their normals.

Therefore,

$$\cos\theta=\frac{|7(2)+(-5)(2)+1(1)|}{\sqrt{75}\sqrt{9}}$$

$$=\frac{5}{15\sqrt{3}}$$

$$=\frac{1}{3\sqrt{3}}$$

Hence,

$$\theta=\cos^{-1}\left(\frac{1}{3\sqrt{3}}\right)$$

Therefore Statement 4 is TRUE.

Hence, the TRUE statements are

$$\boxed{\text{Statements 1 and 4}}$$