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

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Three Dimensional Geometry

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

△ ABC

\frac{9 2}{2}

Answer: (B)

Let

A (x, y, 0)

be equidistant from

P (0, 3, 2), Q (2, 0, 3), R (0, 0, 1)

A P^{2} = A R^{2}

x^{2} + (y - 3)^{2} + 4 = x^{2} + y^{2} + 1 \Rightarrow - 6 y + 13 = 1 \Rightarrow y = 2

A Q^{2} = A R^{2}

(x - 2)^{2} + y^{2} + 9 = x^{2} + y^{2} + 1 \Rightarrow - 4 x + 13 = 1 \Rightarrow x = 3

A = (3, 2, 0)

. With

B = (1, 4, - 1)

and

C = (2, 0, - 2)

A B = (3 - 1)^{2} + (2 - 4)^{2} + (0 + 1)^{2} = 9 = 3

A C = (3 - 2)^{2} + (2 - 0)^{2} + (0 + 2)^{2} = 9 = 3

B C = (1 - 2)^{2} + (4 - 0)^{2} + (- 1 + 2)^{2} = 18 = 32

Since

A B = A C

, the triangle is isosceles; and

A B^{2} + A C^{2} = 9 + 9 = 18 = B C^{2}

, so the angle at

A

9 0^{\circ}

— the triangle is isosceles right-angled, confirming (S1).

Area

= \frac{1}{2} \cdot A B \cdot A C = \frac{1}{2} (3) (3) = \frac{9}{2}

, not

\frac{9 2}{2}

, so (S2) is false.

Only (S1) is true.

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option