Let P be the image of the point Q(7,-2,5)$Q(7,-2,5)$ in the line L: fracx-12=fracy+13=fracz4$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and R(5,p,q)$R(5,p,q)$ be a point on L. Then the square of the area of triangle PQR$\triangle PQR$ is \_\_\_\_. [cite: 3405, 3406]

### Related Formula 1. Area of a \triangle with perpendicular height h$h$ and base b$b$: textArea = frac12 times b times h$$\text{Area} = \frac{1}{2} \times b \times h$$ 2. Since P$P$ is the reflection image of Q$Q$ across line L$L$, the line acts as a perpendicular bisector. For any point R$R$ lying on the line, the height from R$R$ to the line is RT$RT$, and the base QP = 2QT$QP = 2QT$. ### Core Logic Determine parameters for point R$R$ lying directly on line L [cite: 3405, 4071]: frac5-12 = fracp+13 = fracq4 Rightarrow 2 = fracp+13 = fracq4$$\frac{5-1}{2} = \frac{p+1}{3} = \frac{q}{4} \Rightarrow 2 = \frac{p+1}{3} = \frac{q}{4}$$ p+1 = 6 Rightarrow p = 5, quad q = 8 Rightarrow R = (5, 5, 8)$$p+1 = 6 \Rightarrow p = 5, \quad q = 8 \Rightarrow R = (5, 5, 8)$$ [cite: 4071, 4073]

### Step 1: Locate Foot of Perpendicular (T$T$) Let the foot of the perpendicular from Q(7, -2, 5)$Q(7, -2, 5)$ on line L$L$ be T(2lambda+1, 3lambda-1, 4lambda)$T(2\lambda+1, 3\lambda-1, 4\lambda)$ . The directional direction of L$L$ is vecb = 2hati + 3hatj + 4hatk$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ . Vector overrightarrowQT = (2lambda - 6)hati + (3lambda + 1)hatj + (4lambda - 5)hatk$\overrightarrow{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k}$ . Apply orthogonality condition overrightarrowQT cdot vecb = 0$\overrightarrow{QT} \cdot \vec{b} = 0$ : 2(2lambda - 6) + 3(3lambda + 1) + 4(4lambda - 5) = 0$$2(2\lambda - 6) + 3(3\lambda + 1) + 4(4\lambda - 5) = 0$$ 4lambda - 12 + 9lambda + 3 + 16lambda - 20 = 0 Rightarrow 29lambda - 29 = 0 Rightarrow lambda = 1$$4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \Rightarrow 29\lambda - 29 = 0 \Rightarrow \lambda = 1$$ [cite: 4077, 4078] Thus, T = (3, 2, 4)$T = (3, 2, 4)$. ### Step 2: Measure Geometric Distances Compute length QT$QT$ using distance metrics : QT = sqrt(3-7)^2 + (2 - (-2))^2 + (4-5)^2 = sqrt16 + 16 + 1 = sqrt33$$QT = \sqrt{(3-7)^2 + (2 - (-2))^2 + (4-5)^2} = \sqrt{16 + 16 + 1} = \sqrt{33}$$ Since P$P$ is the symmetrical image, base QP = 2QT = 2sqrt33$QP = 2QT = 2\sqrt{33}$. Compute length RT$RT$ representing height from vertex R(5, 5, 8)$R(5, 5, 8)$ to base line at T(3, 2, 4)$T(3, 2, 4)$ : RT = sqrt(5-3)^2 + (5-2)^2 + (8-4)^2 = sqrt4 + 9 + 16 = sqrt29$$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$$ ### Step 3: Calculate Squared Area Compute the \triangle area squared value [cite: 3406, 4081]: textArea = frac12 times QP times RT = frac12 times left(2sqrt33right) times sqrt29 = sqrt957$$\text{Area} = \frac{1}{2} \times QP \times RT = \frac{1}{2} \times \left(2\sqrt{33}\right) \times \sqrt{29} = \sqrt{957}$$ left(textArearight)^2 = 957$$\left(\text{Area}\right)^2 = 957$$ [cite: 4081, 4083] ### Pattern Recognition Because the image geometry creates an isosceles pairing from any point on the mirror line to the object and image point, the area reduces beautifully to 2 times textArea(triangle QTR) = QT times RT$2 \times \text{Area}(\triangle QTR) = QT \times RT$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry