The equation of the plane, parallel to the plane 3x + 4y - 12z =3 and passes through (1, 1, -1), is
Answer & explanation
Correct answer: option 3
Equation of plane parallel to 3x + 4y −12z = 0 is given by,
$3x + 4y −12z+λ=0$
It passes through (1, 2, 3)
$⇒3(1)+4(1)−12(1)+λ=0⇒3+4+λ=0$
$λ=0-19⇒λ=-19$
∴ required plane is
$3x + 4y −12z -19= 0$
$⇒3x + 4y −12z =19$