If for some α and β in R , the intersection of the following three planes x + 4y - 2z = 1 x + 7y - 5z = β x + 5y + = 5 is a line in R ^ 3 , then α + β is equal to:

Correct answer: B. 10 ∆ = 0 ⇒ | cases 1 & 4 & - 2 1 & 7 & - 5 1 & 5 & α cases | = 0 (7α + 25) - (4α + 10) + (- 20 + 14) = 0 3α + 9 = 0 ⇒ α = - 3 Also D _ z = 0 ⇒ | cases 1 & 4 & 1 1 & 7 & β 1 & 5 & 5 cases | = 0 1(35 - 5β) - (15) + 1(4β - 7) = 0 β = 13

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Determinants

If for some α and β in R , the intersection of the following three...

If for some

\alpha

and

\beta

R

, the intersection of the following three planes

x + 4y - 2z = 1

x + 7y - 5z = \beta

x + 5y + \alphaz = 5

is a line in

{R}^{3}

, then

\alpha + \beta

is equal to:

Options

A $0$
B $10$
C $2$
D $- 10$

Correct answer

B. $10$

Step-by-step solution

∆ = 0 ⇒ \left|

\begin{cases}1 & 4 & - 2 \\ 1 & 7 & - 5 \\ 1 & 5 & \alpha\end{cases}

\right| = 0

<br/>\left(7\alpha + 25\right) - \left(4\alpha + 10\right) + \left(- 20 + 14\right) = 0

3\alpha + 9 = 0 ⇒ \alpha = - 3

Also

{D}_{z} = 0 ⇒ \left|

\begin{cases}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5\end{cases}

\right| = 0

1\left(35 - 5\beta\right) - \left(15\right) + 1\left(4\beta - 7\right) = 0

\beta = 13

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