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Negative determinant is positive?Hello, I am looking at this question. "Determine a general form of the equation of the plane through each of the following sets of three points" P1(1,-2,3) P2(1,0,2) P3(-1,4,6) This seems simple enough. I get this matrix https://i.imgur.com/rehV9zn.png Figuring out the determinants for each will get me 12,-2,4 So putting them in point normal form SHOULD net me 12(x-1)-2(y+2)+4(z-3) And the general form of that would be 12x-12-2y-4+4z-12 Which becomes 12x-2y+4z-28 = 0 Simplified to 6x-y+2z-14 = 0 However this answer is WRONG. The actual answer is 6x+y+2z-10 = 0 The mistake I made is that the point normal form should be 12(x-1)+2(y+2)+4(z-3) But I don't understand why. The determinant of the matrix 0 -1 -2 3 Is unquestionably -2. So why is the point normal form using +2 instead? I am especially confused because I have done other questions like this and have gotten answer correct without having to switch around signs. So what is happening here that I'm not seeing? Help would be much appreciated, thank you. |

I admit I don't really follow what you've done. The way I would attack this is form $v_1 = p_2-p_1,~v_2 = p_3-p_1$ $v_1 = (0,2,-1),~v_2 = (-2,6,3)$ $n = v_1 \times v_2 = (12,2,4)$ $n$ is the normal vector to the plane Thus, using $p_1$ as a point on the plane, the scalar equation of the plane is given by $n\cdot ((x,y,z)-p_1) = 0$ or $12(x-1)+2(y+2) + 4(z-3)=0$ or $12x + 2y + 4z = 20$ which reduces to $6x + y + 2z = 10$ |

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For the benefit of others, your method is briefly described here. |

Your 2-by-2 determinant computation is correct as you noticed but your 3-by-3 determinant is where you are making the mistake. Namely, as you compute minors for the matrix they should alternate in sign. The second minor is -2 as you noticed but should gain a minus sign from the determinant computation of the 3-by-3 matrix. Read here for a reminder on how to compute determinants. https://en.wikipedia.org/wiki/Determinant |

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