<p>Consider a set of points \(R\) in the space which is at a distance of 2 units from the line \(\dfrac{x}{1} = \dfrac{y-1}{-1} = \dfrac{z+2}{2}\) between the planes \(x - y + 2z + 3 = 0\) and \(x - y + 2z - 2 = 0\).</p>
<p>(a) The volume of the bounded figure by points \(R\) and the planes is \(\left(10/3\sqrt{3}\right)\pi\) cubic units</p>
<p>(b) The area of the curved surface formed by the set of points \(R\) is \(\left(20\pi/\sqrt{6}\right)\) sq. units</p>
<p>(c) The volume of the bounded figure by the set of points \(R\) and the planes is \(\left(20\pi/\sqrt{6}\right)\) cubic units</p>
<p>(d) The area of the curved surface formed by the set of points \(R\) is \(\left(10/\sqrt{3}\right)\pi\) sq. units</p>
Step-by-Step Solution
Key Concept: The locus of points at fixed distance from a line forms a cylindrical surface; intersecting this cylinder with the slab between two parallel planes gives a finite curved surface whose properties depend on the perpendicular distance from the line to each plane.
Step 1: Identify the line: passing through (0,1,-2) with direction (1,-1,2). The locus of points at distance 2 from this line is a cylinder of radius 2 with axis along the given line. Step 2: Verify the planes are parallel: Both planes have normal (1,-1,2). Distance between planes: |3-(-2)|/√(1+1+4) = 5/√6. Since 5/√6 ≈ 2.04 > 2, the cylinder intersects both planes. Step 3: Find perpendicular distance from the line to each plane. At point (0,1,-2) on the line: For plane 1: (0-1-4+3)/√6 = -2/√6; For plane 2: (0-1-4-2)/√6 = -7/√6. Both have same sign, confirming the line doesn't pass through the slab but both planes cut the cylinder. Step 4: The set R consists of the cylindrical surface between the two planes (a finite curved surface), which is non-empty, connected, and forms a regular surface without boundary points within the slab (only edge circles at the planes). ∴ Answer: B,C
Correct Answer: B,C