Question: A museum curator is restoring a triangular panel from a historical instrument, where the triangle has vertices at $A = (1, 3, -2)$, $B = (4, 1, 1)$, and $C = (2, -1, 3)$. Find the point $P$ on the line through $A$ and $B$ that is closest to point $C$.

["Query: What is the point $P$ on the line through points $A = (1, 3, -2)$ and $B = (4, 1, 1)$ that is closest to point $C = (2, -1, 3)$?", "Restoring historical artifacts often involves precise mathematical modeling—especially when reconstructing intricate parts like triangular panels from ancient instruments. One such task is determining the closest geometric point on a line to a given vertex, such as finding the projection of a vertex onto a face’s edge. In this case, the goal is to find the point $P$ on the line segment through $A$ and $B$ that minimizes the distance to point $C$. This projection is essential in accurately reconstructing the original shape with geometric fidelity.", "---", "### Step-by-Step Solution", "We want to find the point $P$ on the line through $A$ and $B$ that is closest to $C$. This is the orthogonal projection of vector $\vec{OC}$ onto the line defined by vectors $\vec{AB}$ and point $A$.", "#### 1. Define Vector Directions", "Let:\n- $ \vec{A} = (1, 3, -2) $\n- $ \vec{B} = (4, 1, 1) $\n- $ \vec{C} = (2, -1, 3) $", "Compute the direction vector of the line:\n$$\n\vec{d} = \vec{B} - \vec{A} = (4 - 1, 1 - 3, 1 - (-2)) = (3, -2, 3)\n$$", "A point on the line is $A$, so the vector form of the line is:\n$$\n\vec{r}(t) = \vec{A} + t\vec{d} = (1, 3, -2) + t(3, -2, 3), \quad t \in \mathbb{R}\n$$", "The point $P$ is the projection of $C$ onto this line.", "#### 2. Use Orthogonal Projection Formula", "The scalar parameter $t$ for the projection is given by:\n$$\nt = \frac{(\vec{C} - \vec{A}) \cdot \vec{d}}{|\vec{d}|^2}\n$$", "First, compute $ \vec{C} - \vec{A} $:\n$$\n\vec{C} - \vec{A} = (2 - 1, -1 - 3, 3 - (-2)) = (1, -4, 5)\n$$", "Dot product with $ \vec{d} = (3, -2, 3) $:\n$$\n(1, -4, 5) \cdot (3, -2, 3) = 1 \cdot 3 + (-4) \cdot (-2) + 5 \cdot 3 = 3 + 8 + 15 = 26\n$$", "Magnitude squared of $ \vec{d} $:\n$$\n|\vec{d}|^2 = 3^2 + (-2)^2 + 3^2 = 9 + 4 + 9 = 22\n$$", "So,\n$$\nt = \frac{26}{22} = \frac{13}{11}\n$$", "#### 3. Compute Coordinates of $P$", "$$\nP = A + t\vec{d} = (1, 3, -2) + \frac{13}{11}(3, -2, 3)\n$$", "Break into components:\n- $x = 1 + \frac{13}{11} \cdot 3 = 1 + \frac{39}{11} = \frac{11 + 39}{11} = \frac{50}{11}$\n- $y = 3 + \frac{13}{11} \cdot (-2) = 3 - \frac{26}{11} = \frac{33 - 26}{11} = \frac{7}{11}$\n- $z = -2 + \frac{13}{11} \cdot 3 = -2 + \frac{39}{11} = \frac{-22 + 39}{11} = \frac{17}{11}$", "---", "### Final Answer", "The point $P$ on the line through $A$ and $B$ closest to $C$ is:\n$$\n\boxed{\left( \frac{50}{11}, \frac{7}{11}, \frac{17}{11} \right)}\n$$", "This precise geometric construction enables historically accurate facial restoration in museum curation, ensuring that each triangle’s position reflects its true spatial role in the instrument’s design.", "---", "#### Why This Matters in Museum Restoration\nAccurately computing projections ensures that reconstructed components align spatially and aesthetically with the original artifact. By applying linear algebra and vector geometry, curators can digitally and physically reassemble historical instruments with high fidelity, preserving both their form and function."]