Let the co-ordinates of one vertex of \(\triangle \mathrm{ABC}\) be \(\mathrm{A}(0,2, \alpha)\) and the other two vertices lie on the line \(\frac{x+\alpha}{5}=\frac{y-1}{2}=\frac{z+4}{3}\). For \(\alpha \in \mathbb{Z}\), if the area of \(\triangle \mathrm{ABC}\) is 21 sq. units and the line segment BC has length \(2 \sqrt{21}\) units, then \(\alpha^{2}\) is equal to ______.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Three Dimensional Geometry

Let the co-ordinates of one vertex of

△ ABC

A (0, 2, α)

and the other two vertices lie on the line

\frac{x + α}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}

. For

α \in Z

, if the area of

△ ABC

is 21 sq. units and the line segment BC has length

221

units, then

α^{2}

is equal to ______.

Answer: 9

The base

B C

lies along the line with direction

d = (5, 2, 3)

through

P (- α, 1, - 4)

, and

\frac{1}{2} B C \cdot h = 21

with

B C = 221

gives the height

h = 21

. With

A P = (- α, - 1, - α - 4)

, the cross product

A P \times d = (2 α + 5, - (2 α + 20), 2 α - 5)

and

∣ d ∣ = 38

, so

h = \frac{( 2 α + 5 ) ^{2} + ( 2 α + 20 ) ^{2} + ( 2 α - 5 ) ^{2}}{38} = 21

. Squaring gives

12 α^{2} + 80 α + 450 = 798

, i.e.

3 α^{2} + 20 α - 87 = 0

, whose integer root is

α = 3

. Hence

α^{2} = 9

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer