JEE Mathematics 2025 Question with Solution

Given: Point $A$ lies in the $xy$-plane, so $z = 0$. Also, $A(x,y,0)$ is equidistant from $(0,3,2)$, $(2,0,3)$ and $(0,0,1)$. Points $B(1,4,-1)$ and $C(2,0,-2)$ are given.

Find: Whether $ABC$ is an isosceles right angled triangle and whether its area is $\frac{9\sqrt{2}}{2}$.

Since $A$ is equidistant from the three points,

$$AP_1^2 = AP_2^2 = AP_3^2$$

Using $AP_1^2 = AP_2^2$,

$$x^2 + (y-3)^2 + (0-2)^2 = (x-2)^2 + y^2 + (0-3)^2$$ $$x^2 + y^2 - 6y + 13 = x^2 + y^2 - 4x + 13$$ $$-6y = -4x \Rightarrow y = \frac{2x}{3}$$

Using $AP_1^2 = AP_3^2$,

$$x^2 + (y-3)^2 + 4 = x^2 + y^2 + 1$$ $$x^2 + y^2 - 6y + 13 = x^2 + y^2 + 1$$ $$-6y = -12 \Rightarrow y = 2$$

Hence,

$$x = 3$$

So,

$$A = (3,2,0)$$

Now compute the side lengths:

$$AB^2 = (3-1)^2 + (2-4)^2 + (0+1)^2 = 4+4+1 = 9$$ $$AB = 3$$ $$BC^2 = (1-2)^2 + (4-0)^2 + (-1+2)^2 = 1+16+1 = 18$$ $$BC = 3\sqrt{2}$$ $$CA^2 = (2-3)^2 + (0-2)^2 + (-2-0)^2 = 1+4+4 = 9$$ $$CA = 3$$

Thus, $AB = AC$, so the triangle is isosceles. Also,

$$AB^2 + AC^2 = 9 + 9 = 18 = BC^2$$

Hence $\angle A = 90^\circ$, so $ABC$ is an isosceles right angled triangle. Therefore, $S1$ is true.

For area,

$$\text{Area} = \frac{1}{2}(AB)(AC) = \frac{1}{2}\cdot 3 \cdot 3 = \frac{9}{2}$$

So the area is $\frac{9}{2}$, not $\frac{9\sqrt{2}}{2}$. Therefore, $S2$ is false.

The solution concludes option D, but its own working gives $S1$ true and $S2$ false. This matches option A among the given options.

Therefore, the correct option based on the working is A.

Given: The page contains contradictory solution text. One approach incorrectly takes $A=(1,1,1)$, which violates the condition that $A$ lies in the $xy$-plane. The second approach uses $z=0$, which is consistent with the question.

Find: The answer supported by the valid working.

Because $A$ is in the $xy$-plane,

$$z = 0$$

So $A=(x,y,0)$.

Equating squared distances from $A$ to $(0,3,2)$ and $(2,0,3)$ gives

$$x^2 + (y-3)^2 + 4 = (x-2)^2 + y^2 + 9$$

which simplifies to

$$y = \frac{2x}{3}$$

Equating squared distances from $A$ to $(0,3,2)$ and $(0,0,1)$ gives

$$x^2 + (y-3)^2 + 4 = x^2 + y^2 + 1$$

which simplifies to

$$y = 2$$

Hence,

$$x = 3$$

and therefore

$$A=(3,2,0)$$

Now,

$$AB=3, \quad AC=3, \quad BC=3\sqrt{2}$$

So the triangle is isosceles and right angled at $A$. Thus $S1$ is true.

Its area is

$$\frac{1}{2}\times 3 \times 3 = \frac{9}{2}$$

Hence $S2$ is false.

Therefore the defensible answer from the valid solution working is only $S1$ is true, i.e. option A.