Solution: Let the three given points be $ A = (1,1,1) $,

["# Solution: Using Three Given Points in 3D Space — The Foundation of Spatial Geometry", "Understanding how to work with three given points in three-dimensional space is essential in fields ranging from computer graphics and geographic mapping to robotics and scientific visualization. Whether you're calculating a plane, finding the best fit surface, or setting up coordinate systems, making sense of these three points forms the cornerstone of 3D geometry and spatial analysis.", "## Why Three Points Matter in 3D Geometry", "In three-dimensional space, any single point defines a location but not direction or orientation. A single point gives you one coordinate: $ A = (1,1,1) $, but without additional constraints, you cannot define a unique geometric structure. However, when combined with two additional points—say $ B $ and $ C $—three points provide enough information to establish a plane and determine critical elements like normal vectors, distance from origin, and more.", "---", "## Step 1: Define the Points", "Let us define the three known points:", "- $ A = (1, 1, 1) $\n- $ B = (x_2, y_2, z_2) $\n- $ C = (x_3, y_3, z_3) $", "For practical application, suppose $ B = (2, 3, 0) $ and $ C = (0, 2, 2) $ — illustrative points from a typical geometric problem.", "---", "## Step 2: Compute Two Direction Vectors", "From point $ A $, define vectors to $ B $ and $ C $:", "$$\n\vec{AB} = B - A = (2-1, 3-1, 0-1) = (1, 2, -1)\n$$\n$$\n\vec{AC} = C - A = (0-1, 2-1, 2-1) = (-1, 1, 1)\n$$", "These vectors lie in the plane defined by points $ A $, $ B $, and $ C $.", "---", "## Step 3: Calculate the Normal Vector to the Plane", "The cross product $ \vec{n} = \vec{AB} \ imes \vec{AC} $ yields a vector perpendicular to the plane:", "$$\n\vec{n} = \n\begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\n1 & 2 & -1 \\n-1 & 1 & 1 \\n\end{vmatrix}\n= \mathbf{i}(2 \cdot 1 - (-1) \cdot 1) - \mathbf{j}(1 \cdot 1 - (-1) \cdot (-1)) + \mathbf{k}(1 \cdot 1 - 2 \cdot (-1))\n$$\n$$\n= \mathbf{i}(2 + 1) - \mathbf{j}(1 - 1) + \mathbf{k}(1 + 2) = (3, 0, 3)\n$$", "Thus, the plane’s normal vector is $ \vec{n} = (3, 0, 3) $, or simplified $ (1, 0, 1) $.", "---", "## Step 4: Form the Plane Equation", "Using point-normal form of a plane equation:\n$$\nn_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0\n$$", "Substituting $ \vec{n} = (1, 0, 1) $ and point $ A = (1,1,1) $:", "$$\n1(x - 1) + 0(y - 1) + 1(z - 1) = 0\n\quad \Rightarrow \quad x + z - 2 = 0\n$$", "So, the equation of the plane is $ x + z = 2 $.", "---", "## Step 5: Additional Useful Quantities", "- Centroid: If you're interested in balancing the three points, compute the centroid:\n $$\n G = \left( \frac{1+2+0}{3}, \frac{1+3+2}{3}, \frac{1+0+2}{3} \right) = \left(1, 2, 1\right)\n $$", "- Face Area: The magnitude of $ \vec{AB} \ imes \vec{AC} $ gives twice the area of triangle $ ABC $:\n $$\n |\vec{n}| = \sqrt{3^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}, \quad \ ext{Area} = \frac{3\sqrt{2}}{2}\n $$", "---", "## Summary: Why Letting Three Points Define a System Works", "By fixing three non-collinear points $ A $, $ B $, and $ C $ in 3D space, we:", "- Instantly define a plane via the normal vector from the cross product\n- Derive key geometric properties like distances, areas, centroids\n- Enable efficient implementation of 3D algorithms in CAD, robotics, VR, and more", "Mastering this foundational approach empowers you to solve complex spatial problems with confidence — turning abstract coordinates into actionable geometric insight.", "---", "### Further Reading", "- Vector algebra in 3D\n- Plane construction from three points\n- Cross product applications in computer graphics\n- Line and plane equations in 3D space", "---", "Keywords: 3D geometry, plane from three points, vector cross product, normal vector, point trilateral space, coordinate system, spatial analysis."]