Let the area of the triangle formed by the lines $\mathrm{x}+2=\mathrm{y}-1=\mathrm{z}$, $\frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}$ and $\frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-2}{1}$ be A . Then $\mathrm{A}^{2}$ is equal to ______

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Three Dimensional Geometry

Let the area of the triangle formed by the lines

x + 2 = y - 1 = z

\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}

and

\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}

be A . Then

A^{2}

is equal to ______

Answer: 56

Solution

Solving the lines pairwise gives the vertices

A (- 2, 1, 0)

(from

L_{1} \cap L_{2}

B (3, 0, 1)

(from

L_{2} \cap L_{3}

) and

C (0, 3, 2)

(from

L_{1} \cap L_{3}

). Then

A B = (5, - 1, 1)

and

A C = (2, 2, 2)

, with

A B \times A C = (- 4, - 8, 12)

of magnitude

16 + 64 + 144 = 224 = 414

. The area is

A = \frac{1}{2} \cdot 414 = 214

, so

A^{2} = 56

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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