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Question #41734

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all planes in $\mathbb{R}^3$ passing through the origin, whose normal vector has direction cosines $(l, m, n)$ satisfying $n = \frac{1}{2}$ . If the perpendicular distance from the point $A(\sqrt{3}, \sqrt{3}, -1)$ to any plane in $S$ is $2$ , find the value of $\sum_{P \in S} (l^3 + m^3)$ where $(l, m, n)$ are the direction cosines of the normal to the plane $P$ .

Options

A.
$-\frac{3\sqrt{3}}{4}$
B.
$-\frac{11\sqrt{3}}{18}$
C.
$-\frac{3\sqrt{3}}{8}$
D.
$\frac{5\sqrt{3}}{36}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Distance from point to plane|Direction Cosines constraint|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

The question tests the application of the 3D distance formula alongside the fundamental constraint of direction cosines ( $l^2+m^2+n^2=1$ ).

The primary trap lies in solving the absolute value equation for $l+m$ , which yields two potential values.

Students who blindly apply algebraic identities to find $l^3+m^3$ without verifying the domain of $l$ and $m$ will include extraneous complex roots.

By checking the condition $(l+m)^2 \le 2(l^2+m^2)$ for real numbers, the invalid case is rejected, leaving only the geometrically valid planes.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Distance from point to plane
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free