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The equation of the plane, parallel to the plane 3x + 4y - 12z =3 and passes through (1, 1, -1), is |
3x + 4y - 12z = 0 3x + 4y + 12z = 3 3x + 4y - 12z = 19 3x + 4y - 12z = 22 |
3x + 4y - 12z = 19 |
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 $3x + 4y −12z -19= 0$ $⇒3x + 4y −12z =19$ |
