Let A(x, y, z) be a point in x y-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 △ ABC is (9 √2)/2,

Solution

Maths sir · 5 steps

Pin down A first with two distance equations, then the three side lengths settle both statements at once.

1. Find A. A lies in the xy-plane, so A = (x, y, 0). Equate AR² with AP² and AQ², where R = (0,0,1), P = (0,3,2) and Q = (2,0,3). The x² + y² terms cancel each time.

   x²+y²+1 = x²+(y-3)²+4 ⇒ y = 2 ; x²+y²+1 = (x-2)²+y²+9 ⇒ x = 3

2. Side vectors. With A = (3, 2, 0), we get AB⃗ = (-2, 2, -1) and AC⃗ = (-1, -2, -2). So AB = AC = 3, and BC = |(1, -4, -1)| = 3√2.
3. Check S1. AB⃗ · AC⃗ = 2 - 4 + 2 = 0, so the angle at A is 90°. With AB = AC, the triangle is an isosceles right triangle, and 3² + 3² = (3√2)² confirms it. S1 is true.
4. Check S2. The two legs are AB and AC, so the area is 1/2 × 3 × 3 = 9/2. That is not 9√2/2, so S2 is false.
5. So. Only S1 is true, which is option (c).

The idea to keep. For a right triangle, the area is half the product of the two legs. Find the right angle with a zero dot product first, then use the sides that meet there, never the hypotenuse.

Answer from the NTA final answer key. Solution by PrepWiser.