question_answer

If for some \[\alpha \] and \[\beta \] in R, the intersection of the following three planes

\[x+4y-2z=1\]

\[x+7y-5z=\beta \]

\[x+5y+\alpha z=5\]

is a line in \[{{R}^{3}},\] then \[\alpha +\beta \] is equal to

[JEE MAIN Held on 09-01-2020 Morning]

A) \[-10\]               

B) \[0\]

C) \[2\]                 

D) 10

Correct Answer: D

Solution :

[d]   \[\because \]    Three equations have infinitely many solutions, so \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ \alpha =-\,3 \right]\] Putting the value of \[\alpha \] in (3) \[x+4y-2z=1\] and \[x+5y-3z=5\] On solving \[y=z+4\] and \[x=-2z-15\] Substituting these values in (2) \[x+7y-5z=\beta =-2z-15+7z+28-5z\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\left[ \beta =13 \right]\]