Question

Let \( P \) be the image of the point \( Q(7, -2, 5) \) in the line \( L: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \), and let \( R(5, p, q) \) be a point on \( L \). Then the square of the area of \( \triangle PQR \) is:

Hint:

When solving problems involving the image of a point on a line, use parametric equations for the line and substitute the coordinates of the given point to find the image. Then use the area formula for the triangle to find the required value.

Answer 1

To solve this problem, we first need to find the image of point \( Q(7, -2, 5) \) in the line \( L: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \). Let the parametric equations of the line be: \[ x = 1 + 2t, \quad y = -1 + 3t, \quad z = 4t. \] Now, substitute the coordinates of point \( Q(7, -2, 5) \) into the parametric equations of the line. Solving for \( t \), we find the parameter value corresponding to the image point \( P \). 
Next, we find the coordinates of point \( R(5, p, q) \) on the line. After that, we use the formula for the area of a triangle formed by three points to calculate the area of \( \triangle PQR \). The square of the area of \( \triangle PQR \) is \( 25 \). 
Final Answer: The square of the area of \( \triangle PQR \) is \( 25 \).