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Equation of a plane

Problem 9.179 · medium

Find an equation of the plane through the points \( \displaystyle (3, 1, 3),\ (0, 3, -1),\ (-2, 3, -1) \).

\[ \left[\begin{matrix}-3\\2\\-4\end{matrix}\right] \]
PQ.Reviewed
\[ \left[\begin{matrix}-5\\2\\-4\end{matrix}\right] \]
PR.Reviewed
\[ \left[\begin{matrix}0\\2\\1\end{matrix}\right] \]
A normal vector n = PQ × PR (scaled).Reviewed
\[ 5 \]
n · P gives the constant.Reviewed
The plane is 2*y + z = 5.
Reviewed

Answer \( 2 y + z = 5 \)

Lines: 5 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
4	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
5	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	all three points satisfy the equation

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the normal vector via the cross product of displacement vectors and verifies the plane equation using one of the given points. The final equation is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the normal vector via the cross product of displacement vectors and verifies the plane equation using one of the given points. The final equation is correct.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The normal vector calculated in step 3 is (0, 2, 1), which implies the x-component is 0. However, the points (3, 1, 3), (0, 3, -1), and (-2, 3, -1) do not lie on a plane with a normal vector having a zero x-component (the vector PR has an x-component of -5, so the normal must have a non-zero x-component unless the cross product was computed incorrectly). Specifically, PQ = (-3, 2, -4) and PR = (-5, 2, -4). The cross product PQ x PR is (0, -8, 4) or simplified (0, -2, 1). Wait, let's recompute. PQ = Q-P = (0-3, 3-1, -1-3) = (-3, 2, -4). PR = R-P = (-2-3, 3-1, -1-3) = (-5, 2, -4). Cross product: i(2*(-4) - (-4)*2) - j((-3)*(-4) - (-4)*(-5)) + k((-3)*2 - 2*(-5)) = i(-8+8) - j(12-20) + k(-6+10) = 0i - j(-8) + k(4) = (0, 8, 4). This simplifies to (0, 2, 1). So the normal vector IS (0, 2, 1). The equation is 0(x-3) + 2(y-1) + 1(z-3) = 0 => 2y - 2 + z - 3 = 0 => 2y + z = 5. The solution is actually correct. Why did I think it was wrong? Let me re-read the points. P(3,1,3), Q(0,3,-1), R(-2,3,-1). Note that Q and R have the same y and z coordinates. The line QR is parallel to the x-axis. The vector QR is (-2, 0, 0). The vector PQ is (-3, 2, -4). The normal must be perpendicular to QR, so its x-component must be 0. The calculation is correct. The solution is sound.
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the normal vector and the plane equation. The intermediate steps show the correct cross product result and constant calculation.
gpt-oss:20b: pass 2026-09-29

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/plane_through_points, checked 2026-10-13 with SymPy 1.14.0.