<p>A variable plane at a distance of 1 unit from the origin cuts the coordinate axes at <em>A</em>, <em>B</em> and <em>C</em>. If the centroid <em>D</em>(<em>x</em>, <em>y</em>, <em>z</em>) of triangle <em>ABC</em> satisfies the relation \(\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} = k\), then the value of <em>k</em> is __________.</p>
Step-by-Step Solution
Key Concept: A plane at distance 1 from origin has equation lx + my + nz = 1 where l² + m² + n² = 1. The intercepts A, B, C are at (1/l, 0, 0), (0, 1/m, 0), (0, 0, 1/n), so the centroid is D(1/3l, 1/3m, 1/3n). Substituting into the given relation yields the constant k.
Step 1: Write the equation of plane at distance 1 from origin. For a plane lx + my + nz = 1 where l, m, n are direction cosines, the distance from origin is 1 (automatically satisfied). Condition: l^2 + m^2 + n^2 = 1 Step 2: Find intercepts on coordinate axes. Setting y = z = 0: x = 1/l, so A = (1/l, 0, 0) Setting x = z = 0: y = 1/m, so B = (0, 1/m, 0) Setting x = y = 0: z = 1/n, so C = (0, 0, 1/n) Step 3: Find centroid D of triangle ABC. D = ((1/l + 0 + 0)/3, (0 + 1/m + 0)/3, (0 + 0 + 1/n)/3) D = (1/3l, 1/3m, 1/3n) Therefore: x = 1/3l, y = 1/3m, z = 1/3n Step 4: Calculate 1/x^2 + 1/y^2 + 1/z^2. 1/x^2 = (3l)^2, 1/y^2 = (3m)^2, 1/z^2 = (3n)^2 1/x^2 + 1/y^2 + 1/z^2 = 9l^2 + 9m^2 + 9n^2 = 9(l^2 + m^2 + n^2) = 9(1) = 9 ∴ Answer: k = 9
Correct Answer: 9