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:

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Matrices

If for some

α

and

β

in R, the intersection of the following three planes

x + 4 y - 2 z = 1

x + 7 y - 5 z = β

x + 5 y + α z = 5

is a line in

R^{3}

, then

α + β

is equal to:

Answer: (C)

For the three planes to meet in a line, the coefficient determinant must vanish. Expanding

111475 - 2 - 5 α

by subtracting the first row from the others gives

3 α + 9 = 0

, so

α = - 3

. Consistency for a whole line of solutions then requires

111475 1 β 5 = 0

, which gives

13 - β = 0

, so

β = 13

. Hence

α + β = - 3 + 13 = 10

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option