Let P be the plane such that it contains the straight line

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

and is perpendicular to the plane

$x+2y+3z=4$

Let P₁ 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?

Correct Options:

1. The equation of the plane P is $7x-5y+z=-10$

2. The acute angle between the plane P and the plane $2x+2y+z=3$ is ${\cos }^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Step-by-Step Solution:

Step 1: Find the equation of the plane P

The given line is:

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

This line passes through the point A(1, 3, -2) and has direction ratios parallel to the vector:

$\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$

Plane P contains this line, so plane P passes through (1, 3, -2), and its normal vector $\overrightarrow{n}=a\hat{i}+b\hat{j}+c\hat{k}$ is perpendicular to $\overrightarrow{b}$:

$2a+3b+c=0$

Plane P is also perpendicular to the plane $x+2y+3z=4$, whose normal vector is $\overrightarrow{n_0}=\hat{i}+2\hat{j}+3\hat{k}$. Therefore, $\overrightarrow{n}$ is perpendicular to $\overrightarrow{n_0}$:

$a+2b+3c=0$

The normal vector $\overrightarrow{n}$ can be found by taking the cross product of $\overrightarrow{b}$ and $\overrightarrow{n_0}$:

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 2& 3\end{array}\right|=\hat{i}(9-2)-\hat{j}(6-1)+\hat{k}(4-3)=7\hat{i}-5\hat{j}+\hat{k}$

Now, using the point-normal form, the equation of plane P passing through (1, 3, -2) with normal vector $(7,-5,1)$ is:

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

$7x-7-5y+15+z+2=0$

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

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

Thus, the first statement is TRUE.

Step 2: Calculate the acute angle between plane P and plane $2x+2y+z=3$

Let $\overrightarrow{n_1}=7\hat{i}-5\hat{j}+\hat{k}$ be the normal to plane P, and $\overrightarrow{n_2}=2\hat{i}+2\hat{j}+\hat{k}$ be the normal to the given plane.

The cosine of the acute angle $\theta$ between the two planes is given by:

$\cos \theta =\frac{\left|\overrightarrow{n_1}·\overrightarrow{n_2}\right|}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}$

Evaluating the dot product and magnitudes:

$\overrightarrow{n_1}·\overrightarrow{n_2}=\left(7\right)\left(2\right)+(-5)\left(2\right)+\left(1\right)\left(1\right)=14-10+1=5$

$\left|\overrightarrow{n_1}\right|=\sqrt{7^2+{(-5)}^2+1^2}=\sqrt{49+25+1}=\sqrt{75}=5\sqrt{3}$

$\left|\overrightarrow{n_2}\right|=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$

Substituting these values into the angle formula:

$\cos \theta =\frac{5}{\left(5\sqrt{3}\right)\left(3\right)}=\frac{1}{3\sqrt{3}}$

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

Thus, the fourth statement is also TRUE.