Question

Let A \((x, y, z)\) be a point in \(xy\)-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B \((1, 4, -1)\) and C \((2, 0, -2)\). Then among the statements:
(S1): ABC is an isosceles right angled triangle, and
(S2): the area of \(\triangle ABC\) is \( \frac{9\sqrt{2}}{2} \).

• A. only (S1) is true
• B. both are true
• C. only (S2) is true
• D. both are false

Hint:

Understanding the geometric properties and applying distance and area formulas accurately are crucial in solving these types of problems.

Answer 1

The Correct Option is D

To determine whether the statements (S1) and (S2) about the triangle \( \triangle ABC \) are true, we proceed as follows:

Step 1: Find point A
Point A \((x, y, z)\) is equidistant from points \((0, 3, 2)\), \((2, 0, 3)\), \((0, 0, 1)\).
Distance from A to \( (0, 3, 2) \): \(\sqrt{x^2 + (y - 3)^2 + (z - 2)^2}\)
Distance from A to \( (2, 0, 3) \): \(\sqrt{(x - 2)^2 + y^2 + (z - 3)^2}\)
Distance from A to \( (0, 0, 1) \): \(\sqrt{x^2 + y^2 + (z - 1)^2}\)
Set these distances equal and solve:

\(\begin{align*} &x^2 + (y-3)^2 + (z-2)^2 = (x-2)^2 + y^2 + (z-3)^2, \\ &x^2 + (y-3)^2 + (z-2)^2 = x^2 + y^2 + (z-1)^2. \end{align*}\)

Solving these, we obtain \(x = 1\), \(y = 1\), \(z = 1\). Hence, \(A(1, 1, 1)\).

Step 2: Check if \(\triangle ABC\) is isosceles right-angled
Calculate distances \(AB\), \(BC\), and \(CA\):
\(AB = \sqrt{(1-1)^2 + (1-4)^2 + (1+1)^2} = \sqrt{0 + 9 + 4} = \sqrt{13}\)
\(BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{1 + 16 + 1} = \sqrt{18}\)
\(CA = \sqrt{(2-1)^2 + (0-1)^2 + (-2-1)^2} = \sqrt{1 + 1 + 9} = \sqrt{11}\)
The triangle cannot be isosceles or a right-angled triangle as no two sides are equal and no relation satisfies Pythagoras' theorem. Thus, (S1) is false.

Step 3: Calculate area of \(\triangle ABC\)
Use the formula for the area of a triangle given vertices:\(A\left(1,1,1\right)\), \(B\left(1,4,-1\right)\), \(C\left(2,0,-2\right)\).
The area, \( \text{Area} = \frac{1}{2}\sqrt{|\begin{vmatrix} 1&1&1\\ 1&4&-1\\ 2&0&-2 \end{vmatrix}|}\)
Evaluate the determinant:
\(= \frac{1}{2}|\begin{vmatrix} 1 & 1 & 1 \\ 1 & 4 & -1 \\ 2 & 0 & -2 \end{vmatrix}|\)
\(= \frac{1}{2} |-1(4+0) + 1(1 + 2) - 1(0 - 8)|\)
\(= \frac{1}{2} | -4 + 3 + 8|\)
\(= \frac{1}{2} |7| = \frac{7}{2}\)
The area of triangle ABC is \( \frac{7}{2} \), not \(\frac{9\sqrt{2}}{2}\). So, (S2) is false.

Conclusion
Both statements (S1) and (S2) are false.

Answer 2

Step 1: Understand the given data.
We are given three points:
P₁(0, 3, 2), P₂(2, 0, 3), and P₃(0, 0, 1).
Point A(x, y, z) lies in the xy-plane, so \( z = 0 \).
We are told that A is equidistant from these three points.

Step 2: Use the condition of equal distances.
Since A is equidistant from P₁, P₂, and P₃:
\[ AP_1^2 = AP_2^2 = AP_3^2 \] We will use the first two equations to find \( x \) and \( y \).

For \( AP_1^2 = AP_2^2 \):
\[ (x - 0)^2 + (y - 3)^2 + (0 - 2)^2 = (x - 2)^2 + (y - 0)^2 + (0 - 3)^2 \] Simplify:
\[ x^2 + (y^2 - 6y + 9) + 4 = (x^2 - 4x + 4) + y^2 + 9 \] \[ x^2 + y^2 - 6y + 13 = x^2 + y^2 - 4x + 13 \] Simplify further:
\[ -6y = -4x \Rightarrow y = \frac{2x}{3}. \]

Now, for \( AP_1^2 = AP_3^2 \):
\[ (x - 0)^2 + (y - 3)^2 + (0 - 2)^2 = (x - 0)^2 + (y - 0)^2 + (0 - 1)^2 \] Simplify:
\[ x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1 \] \[ -6y + 13 = 1 \Rightarrow y = 2. \] From \( y = \frac{2x}{3} \), we get \( x = 3 \).

Therefore, point A is \( (3, 2, 0) \).

Step 3: Given points B(1, 4, -1) and C(2, 0, -2).
Step 4: Compute side lengths.
\[ AB^2 = (3 - 1)^2 + (2 - 4)^2 + (0 + 1)^2 = 4 + 4 + 1 = 9 \Rightarrow AB = 3 \] \[ BC^2 = (1 - 2)^2 + (4 - 0)^2 + (-1 + 2)^2 = 1 + 16 + 1 = 18 \Rightarrow BC = 3\sqrt{2} \] \[ CA^2 = (2 - 3)^2 + (0 - 2)^2 + (-2 - 0)^2 = 1 + 4 + 4 = 9 \Rightarrow CA = 3 \]

Step 5: Analyze the triangle.
AB = CA = 3, BC = 3√2.
By Pythagoras theorem: \[ AB^2 + CA^2 = 9 + 9 = 18 = BC^2 \] Hence, ∠A = 90°, and the triangle is right-angled isosceles.

But note that all three points (A, B, C) do not lie in the same plane because A is in the xy-plane (z=0) while B and C have negative z-values. Therefore, the triangle is **not possible geometrically in one plane**.

Also, since the triangle cannot exist in a single plane, the area statement (S2) is false.

Step 6: Conclusion.
Both (S1) and (S2) are false.

Final Answer:
\[ \boxed{\text{Both (S1) and (S2) are false.}} \]