Consider the vectors \[ \vec{x}=\hat{i}+2 \hat{j}+3 \hat{k}, \quad \vec{y}=2 \hat{i}+3 \hat{j}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{i}+\hat{j}+2 \hat{k} \] For two distinct positive real numbers \(\alpha\) and \(\beta\), define \[ \vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y} . \] If the vectors \(\vec{X}, \vec{Y}\), and \(\vec{Z}\) lie in a plane, the value of \(\alpha+\beta-3\) is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Vector Algebra

Consider the vectors

x = \hat{i} + 2 \hat{j} + 3 \hat{k}, y = 2 \hat{i} + 3 \hat{j} + \hat{k}, and z = 3 \hat{i} + \hat{j} + 2 \hat{k}

For two distinct positive real numbers

α

and

β

, define

X = α x + β y - z, Y = α y + β z - x, and Z = α z + β x - y .

If the vectors

X, Y

, and

Z

lie in a plane, the value of

α + β - 3

is ____.

Answer: -2

The vectors are coplanar when their scalar triple product vanishes. Expressing

X, Y, Z

through

x, y, z

, that product equals the determinant

α - 1 β β α - 1 - 1 β α

times

[x y z]

. Since

[x y z] \neq = 0

, the determinant must be zero:

α^{3} + β^{3} - 1 + 3 α β = 0

. This is

α^{3} + β^{3} + (- 1)^{3} - 3 (α) (β) (- 1) = 0

, which factors as

(α + β - 1) (α^{2} + β^{2} + 1 - α β + α + β) = 0

. For distinct positive

α, β

the second factor is positive, so

α + β = 1

and

α + β - 3 = - 2

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer