The equation of the plane containing the line of intersection of the planes x+y+z-6=0 and 2x+3y+4y+5=0 and passing through the point (1, 1, 1), is
Answer & explanation
Correct answer: option 1
The equation of the plane through the line of intersection of the given planes is
$(x+y+z-6) + λ(2x + 3y+4z +5)=0$
If it passes through (1, 1, 1), we have
$-3+14λ=0⇒λ=\frac{3}{14}$
Putting $λ =\frac{3}{14}$ in (i), we obtain the equation of the required plane as $20x + 23y + 26z - 69= 0 $, is