If for some α and β in R, intersection of following three planes x+4 y-2 z=1 ; x+7 y-5 z=β

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3 and 4 .Determinants and Matrices

hard

If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to

$10$

$-10$

$2$

$0$

(JEE MAIN-2020)

Solution

For planes to intersect on a line

$\Rightarrow$ there should be infinite solution of the given system of equations for infinite solutions

$\Delta=\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0 \Rightarrow 3 \alpha+9=0 \Rightarrow \alpha=-3$

$\Delta_{z}=\left|\begin{array}{ccc}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0 \Rightarrow 13-\beta=0 \Rightarrow \beta=13$

Also for $\alpha=-3$ and $b=13 \Delta_{x}=\Delta_{y}=0$

$\alpha+\beta=-3+13=10$

Standard 12

Mathematics

If $\alpha+\beta+\gamma=2 \pi$, then the system of equations

$x+(\cos \gamma) y+(\cos \beta) z=0$

$(\cos \gamma) x+y+(\cos \alpha) z=0$

$(\cos \beta) x+(\cos \alpha) y+z=0$

has :

hard

(JEE MAIN-2021)

The system of equations $\begin{array}{l}\alpha x + y + z = \alpha – 1\\x + \alpha y + z = \alpha – 1\\x + y + \alpha z = \alpha – 1\end{array}$ has no solution, if $\alpha $ is

hard

(AIEEE-2005)

If $\left| {\begin{array}{*{20}{c}}{x – 4}&{2x}&{2x}\\{2x}&{x – 4}&{2x}\\{2x}&{2x}&{x – 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x – A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .

medium

(JEE MAIN-2018)

A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 – x}&{ – 6}&3\\{ – 6}&{3 – x}&3\\3&3&{ – 6 – x}\end{array}\,} \right| = 0$ is

easy

For the system of linear equations

$2 x+4 y+2 a z=b$

$x+2 y+3 z=4$

$2 x-5 y+2 z=8$

which of the following is NOT correct?

hard

(JEE MAIN-2023)

If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to

Solution

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If $\alpha+\beta+\gamma=2 \pi$, then the system of equations

$x+(\cos \gamma) y+(\cos \beta) z=0$

$(\cos \gamma) x+y+(\cos \alpha) z=0$

$(\cos \beta) x+(\cos \alpha) y+z=0$

has :

For the system of linear equations

$2 x+4 y+2 a z=b$

$x+2 y+3 z=4$

$2 x-5 y+2 z=8$

which of the following is NOT correct?