General Equation of a Plane from Normal and Point

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

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

1. Direction Vector of $$L_1$$
Line $$L_1$$ is parallel to the vector $$\vec{v}_1 = 2\hat{i} - \hat{j} + 3\hat{k}$$. It passes through point $$P(1, 2, 1)$$.

2. Direction Vector of $$L_2$$
Line $$L_2$$ is the intersection of planes $$x + 2y - z = 3$$ and $$2x - y + 3z = 1$$. Its direction vector $$\vec{v}_2$$ is perpendicular to both normal vectors $$\vec{n}_1 = (1, 2, -1)$$ and $$\vec{n}_2 = (2, -1, 3)$$:
$$\vec{v}_2 = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix} = 5\hat{i} - 5\hat{j} - 5\hat{k}$$
We can simplify this parallel vector to $$(1, -1, -1)$$.

3. Equation of Plane $$\pi$$
Plane $$\pi$$ contains $$L_1$$ and is parallel to $$L_2$$. Its normal vector $$\vec{n}_{\pi}$$ is:
$$\vec{n}_{\pi} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & -1 & -1 \end{vmatrix} = 4\hat{i} + 5\hat{j} - \hat{k}$$
Since the plane contains point $$P(1, 2, 1)$$, its equation is:
$$4(x - 1) + 5(y - 2) - 1(z - 1) = 0 \implies 4x + 5y - z - 13 = 0$$

4. Distance from $$Q(2, 3, 4)$$ to Plane $$\pi$$
Using the point-to-plane distance formula:
$$\text{Distance} = \frac{|4(2) + 5(3) - (4) - 13|}{\sqrt{4^2 + 5^2 + (-1)^2}} = \frac{|8 + 15 - 4 - 13|}{\sqrt{16 + 25 + 1}} = \frac{6}{\sqrt{42}}$$
Note: $$\frac{6}{\sqrt{42}} = \frac{\sqrt{36}}{\sqrt{42}} = \sqrt{\frac{6}{7}}$$.

For plane $$P$$ to contain $$L_1$$ and be at a maximum distance from parallel line configurations, it must be parallel to $$L_2$$.  

The normal vector $$\vec{n}$$ to the plane must be perpendicular to both $$\vec{u}$$ and $$\vec{v}$$.

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(-3 - 1) - \hat{j}(3 - 2) + \hat{k}(1 + 2) = -4\hat{i} - \hat{j} + 3\hat{k}$$
The plane passes through the point $$\vec{a}(1, 2, 3)$$ on $$L_1$$. The vector equation of the plane is:
$$-4(x - 1) - 1(y - 2) + 3(z - 3) = 0 \implies -4x - y + 3z + 4 + 2 - 9 = 0 \implies 4x + y - 3z + 3 = 0$$
Perpendicular distance from the origin $$(0,0,0)$$ to the plane $$4x + y - 3z + 3 = 0$$:
$$\text{Distance} = \frac{|3|}{\sqrt{4^2 + 1^2 + (-3)^2}} = \frac{3}{\sqrt{16 + 1 + 9}} = \frac{3}{\sqrt{26}}$$