<p>A variable plane passes through a fixed point \((3, 2, 1)\) and meets \(x\), \(y\) and \(z\) axes at \(A\), \(B\) and \(C\), respectively. A plane is drawn parallel to \(yz\)-plane through \(A\), a second plane is drawn parallel to \(zx\)-plane through \(B\) and a third plane is drawn parallel to \(xy\)-plane through \(C\). Then the locus of the point of intersection of these three planes, is</p>
<p>\(\dfrac{x}{3} + \dfrac{y}{2} + \dfrac{z}{1} = 1\)</p>
<p>\(x + y + z = 6\)</p>
<p>\(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{11}{6}\)</p>
<p>\(\dfrac{3}{x} + \dfrac{2}{y} + \dfrac{1}{z} = 1\)</p>
Step-by-Step Solution
Key Concept: The intercept form of a plane is x/a + y/b + z/c = 1, where a, b, c are x, y, z intercepts. The three planes parallel to coordinate planes through A(a,0,0), B(0,b,0), C(0,0,c) intersect at point P(a,b,c), which must satisfy the intercept equation since the original plane passes through (3,2,1).
Step 1: Let the variable plane meet the axes at A(a,0,0), B(0,b,0), and C(0,0,c). The equation of this plane in intercept form is: x/a + y/b + z/c = 1 Step 2: Since the plane passes through the fixed point (3,2,1), we have: 3/a + 2/b + 1/c = 1 Step 3: The plane parallel to yz-plane through A(a,0,0) has equation x = a. The plane parallel to zx-plane through B(0,b,0) has equation y = b. The plane parallel to xy-plane through C(0,0,c) has equation z = c. Step 4: These three planes intersect at the point P(a,b,c). As the variable plane changes, the point P traces a locus. Step 5: From Step 2, we have the constraint: 3/a + 2/b + 1/c = 1. If we denote the coordinates of point P as (x,y,z), then a = x, b = y, c = z. Step 6: Substituting into the constraint: 3/x + 2/y + 1/z = 1 ∴ Answer: The locus is 3/x + 2/y + 1/z = 1 (Option D)
Correct Answer: D