1. Key Concepts and Formulas

• Foot of the Perpendicular from a Point to a Line: To find the shortest distance from an external point $P$ to a line $L$, we find the point $T$ on $L$ such that the vector $\vec{PT}$ is perpendicular to the direction vector of the line $L$. This condition is expressed by their dot product being zero: $\vec{PT} \cdot \vec{d} = 0$.
• Properties of an Isosceles Triangle: If two points $Q$ and $R$ on a line $L$ are equidistant from an external point $P$ (i.e., $PQ = PR$), then $\triangle PQR$ is an isosceles triangle. The altitude from $P$ to the base $QR$ (which is $PT$, where $T$ is the foot of the perpendicular from $P$ to $L$) will also bisect the base $QR$. Thus, $T$ is the midpoint of $QR$.
• Pythagorean Theorem: In a right-angled triangle $\triangle PTQ$, where $PT$ is perpendicular to $TQ$, we have $PQ^2 = PT^2 + QT^2$.
• Area of a Triangle: The area of $\triangle PQR$ is given by $\frac{1}{2} \times \text{base} \times \text{height}$. In this case, the base is $QR$ and the height is $PT$.

2. Step-by-Step Solution

Step 1: Parametrize the Line and Identify Key Vectors

We are given the line $L$ in symmetric form: $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$ To represent any general point on this line, we introduce a parameter $t$. This allows us to express the coordinates of any point on $L$ in terms of $t$. Let a general point $T$ on the line $L$ be $(x, y, z)$. By setting each fraction equal to $t$: $\frac{x+1}{2} = t \implies x = 2t - 1$ $\frac{y+2}{3} = t \implies y = 3t - 2$ $\frac{z-1}{2} = t \implies z = 2t + 1$ So, any point on the line $L$ can be represented as $T(2t-1, 3t-2, 2t+1)$.

The given external point is $P(4,2,7)$.

The direction vector of the line $L$, denoted as $\vec{d}$, is derived from the denominators of the symmetric form: $\vec{d} = (2, 3, 2)$

Now, we form the vector $\vec{PT}$ connecting point $P$ to the general point $T$ on the line. This vector is crucial for finding the foot of the perpendicular. $\vec{PT} = ( (2t-1) - 4, (3t-2) - 2, (2t+1) - 7 )$ $\vec{PT} = (2t-5, 3t-4, 2t-6)$

Step 2: Find the Foot of the Perpendicular, $T$

The point $T$ on line $L$ that is the foot of the perpendicular from $P$ is characterized by the fact that the vector $\vec{PT}$ is perpendicular to the line $L$. This means $\vec{PT}$ must be perpendicular to the direction vector $\vec{d}$ of the line. The dot product of two perpendicular vectors is zero. $\vec{PT} \cdot \vec{d} = 0$ Substitute the components of $\vec{PT}$ and $\vec{d}$: $(2t-5)(2) + (3t-4)(3) + (2t-6)(2) = 0$ Expand and simplify the equation to solve for $t$: $(4t - 10) + (9t - 12) + (4t - 12) = 0$ $(4t + 9t + 4t) + (-10 - 12 - 12) = 0$ $17t - 34 = 0$ $17t = 34$ $t = 2$ Now, substitute the value $t=2$ back into the parametric coordinates of $T$ to find the exact coordinates of the foot of the perpendicular: $T_x = 2(2) - 1 = 4 - 1 = 3$ $T_y = 3(2) - 2 = 6 - 2 = 4$ $T_z = 2(2) + 1 = 4 + 1 = 5$ So, the foot of the perpendicular from $P$ to the line $L$ is $T(3,4,5)$.

Step 3: Calculate the Height of the Triangle, $PT$

The distance $PT$ is the perpendicular distance from point $P(4,2,7)$ to the line $L$. As established by the properties of an isosceles triangle, this distance $PT$ is the height of $\triangle PQR$ with respect to the base $QR$. We use the 3D distance formula between $P(4,2,7)$ and $T(3,4,5)$: $PT = \sqrt{(3-4)^2 + (4-2)^2 + (5-7)^2}$ $PT = \sqrt{(-1)^2 + (2)^2 + (-2)^2}$ $PT = \sqrt{1 + 4 + 4}$ $PT = \sqrt{9} = 3$ Thus, the height of $\triangle PQR$ is $PT = 3$ units.

Step 4: Calculate Half the Base of the Triangle, $QT$

We are given that points $Q$ and $R$ are on the line $L$ and are at a distance $\sqrt{26}$ from $P$. This means $PQ = \sqrt{26}$. Consider the right-angled triangle $\triangle PTQ$. The hypotenuse is $PQ$, one leg is $PT$ (the height we just calculated), and the other leg is $QT$. We apply the Pythagorean theorem: $PQ^2 = PT^2 + QT^2$ Substitute the known values $PQ = \sqrt{26}$ and $PT = 3$: $(\sqrt{26})^2 = (3)^2 + QT^2$ $26 = 9 + QT^2$ $QT^2 = 26 - 9$ $QT^2 = 17$ $QT = \sqrt{17}$ Since $T$ is the midpoint of $QR$ (because $\triangle PQR$ is isosceles and $PT$ is the altitude), the full base $QR$ is twice the length of $QT$: $QR = 2 \times QT = 2\sqrt{17}$

Step 5: Calculate the Area of $\triangle PQR$

The area of a triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{height}$. Here, the base is $QR = 2\sqrt{17}$ and the height is $PT = 3$. $\text{Area}(\triangle PQR) = \frac{1}{2} \times QR \times PT$ $\text{Area}(\triangle PQR) = \frac{1}{2} \times (2\sqrt{17}) \times 3$ $\text{Area}(\triangle PQR) = 3\sqrt{17}$

Step 6: Calculate the Square of the Area of $\triangle PQR$

The question asks for the square of the area of $\triangle PQR$. $(\text{Area}(\triangle PQR))^2 = (3\sqrt{17})^2$ $(\text{Area}(\triangle PQR))^2 = 3^2 \times (\sqrt{17})^2$ $(\text{Area}(\triangle PQR))^2 = 9 \times 17$ $(\text{Area}(\triangle PQR))^2 = 153$

3. Common Mistakes & Tips

• Incorrectly Identifying the Triangle Type: Failing to recognize that $\triangle PQR$ is an isosceles triangle can lead to an incorrect approach for finding the base or height.
• Errors in Finding the Foot of the Perpendicular: This step involves algebraic manipulation and dot product calculation. A common mistake is an arithmetic error or incorrectly setting up the dot product equation.
• Misapplication of Pythagorean Theorem: Ensure you correctly identify the hypotenuse and legs in the right-angled triangle $\triangle PTQ$.

4. Summary

We began by parametrizing the given line and identifying its direction vector. We then found the foot of the perpendicular $T$ from point $P$ to the line by utilizing the condition that $\vec{PT}$ is perpendicular to the line's direction vector. This allowed us to calculate the height $PT$ of the triangle. Recognizing $\triangle PQR$ as an isosceles triangle and using the given distance $PQ$ along with the calculated height $PT$ in the right-angled triangle $\triangle PTQ$, we determined half the base $QT$. Finally, we calculated the area of $\triangle PQR$ using the base $QR=2QT$ and height $PT$, and then squared this area to arrive at the final answer.

The final answer is $\boxed{153}$.