Step 1: Conceptual Overview: The problem evaluates fundamental principles of planes and lines in three-dimensional geometry. This includes determining plane equations, calculating the angles between planes, the angle between a line and an axis, and identifying normal vectors.
Step 2: Detailed Analysis:
(A): The general equation for a plane passing through \((x_0, y_0, z_0)\) with a normal vector having direction ratios (a, b, c) is \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\).
Given the point (1, -1, 2) and normal vector (2, 3, 2), the equation is:
\(2(x-1) + 3(y-(-1)) + 2(z-2) = 0\)
\(2x-2+3y+3+2z-4=0\)
\(2x+3y+2z-3=0\)
\(2x+3y+2z=3\)
Statement (A) is confirmed as true.
(B): The angle \(\theta\) between two planes is equivalent to the angle between their normal vectors.
The normal vectors are \(\vec{n_1} = (2, -1, 1)\) and \(\vec{n_2} = (1, 1, 2)\).
The cosine of the angle is calculated as:
\(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|2(1)+(-1)(1)+1(2)|}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}\).
Thus, \(\theta = \cos^{-1}(1/2) = \pi/3\).
Statement (B) is confirmed as true.
(C): The normal vector to the plane is \(\vec{n} = (4, 8, 1)\). The z-axis is represented by the vector \(\vec{k} = (0, 0, 1)\).
The angle \(\alpha\) between \(\vec{n}\) and the z-axis is determined by:
\(\cos\alpha = \frac{|\vec{n} \cdot \vec{k}|}{|\vec{n}||\vec{k}|} = \frac{|4(0)+8(0)+1(1)|}{\sqrt{4^2+8^2+1^2}\sqrt{1}} = \frac{1}{\sqrt{16+64+1}} = \frac{1}{\sqrt{81}} = \frac{1}{9}\).
Therefore, \(\alpha = \cos^{-1}(1/9)\). The statement incorrectly claims \(\sin^{-1}(1/9)\).
Statement (C) is determined to be false.
(D): The normal vector to a plane is parallel to the line connecting two points.
The normal vector is calculated as:
\(\vec{n} = (3-2, 4-(-1), -1-5) = (1, 5, -6)\).
The plane passes through the point (3, -3, 1).
The equation of the plane is:
\(1(x-3) + 5(y-(-3)) - 6(z-1) = 0\)
\(x-3+5y+15-6z+6=0\)
\(x+5y-6z+18=0\)
The provided statement gives \(x+5y+6z=-18\). The sign of the z-term is incorrect.
Statement (D) is determined to be false.
(E): For the plane \(2x - y + 2z = 5\), a normal vector is \(\vec{n} = 2\vec{i} - \vec{j} + 2\vec{k}\). Any scalar multiple of a normal vector is also a normal vector.
The vector \(\tfrac{1}{3}(2\vec{i} - \vec{j} + 2\vec{k})\) is a scalar multiple (specifically, the unit normal vector \(\hat{n}\)) of the original normal vector, and thus it is also a normal vector.
Statement (E) is confirmed as true.
Step 3: Conclusion: The correct statements are (A), (B), and (E). This corresponds to option (D).