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
$$\boxed{\text{Statements 1 and 4}}$$
91- 6303239042
