Let P be a plane passing through the points \((1,0,1)\), \((1,-2,1)\) and \((0,1,-2)\). Let a vector \(\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}\) be such that \(\overrightarrow{\mathrm{a}}\) is parallel to the plane P, perpendicular to \((\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\vec{a} \cdot(\hat{i}+\hat{j}+2 \hat{k})=2\), then \((\alpha-\beta+\gamma)^{2}\) equals ______.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Vector Algebra

Let P be a plane passing through the points

(1, 0, 1)

(1, - 2, 1)

and

(0, 1, - 2)

. Let a vector

a = α \hat{i} + β \hat{j} + γ \hat{k}

be such that

a

is parallel to the plane P, perpendicular to

(\hat{i} + 2 \hat{j} + 3 \hat{k})

and

a \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2

, then

(α - β + γ)^{2}

equals ______.

Answer: 81

First, find the normal to plane P using two vectors in the plane:

v_{1} = (1, - 2, 1) - (1, 0, 1) = (0, - 2, 0)

v_{2} = (0, 1, - 2) - (1, 0, 1) = (- 1, 1, - 3)

Normal

n = v_{1} \times v_{2} = (6, 0, - 2)

or simplified

(3, 0, - 1)

.

Since

a

is parallel to P,

a \cdot n = 0

3 α - γ = 0

⟹

γ = 3 α

.

Since

a ⊥ (\hat{i} + 2 \hat{j} + 3 \hat{k})

α + 2 β + 3 γ = 0

⟹

α + 2 β + 9 α = 0

⟹

β = - 5 α

.

Using

a \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2

α + β + 2 γ = 2

⟹

α - 5 α + 6 α = 2

⟹

2 α = 2

⟹

α = 1

.

Thus

β = - 5

γ = 3

, and

(α - β + γ)^{2} = (1 - (- 5) + 3)^{2} = 9^{2} = 81

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer