Key Concepts and Formulas

1. Parametric Form of a Line: A line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be expressed as: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$ Any point on this line can be represented parametrically as $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$, where $\lambda$ is a scalar parameter.

2. Equation of a Plane: A linear equation of the form $Ax + By + Cz = D$ represents a plane in three-dimensional space.

3. Distance Formula in 3D: The square of the distance $d^2$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

Step-by-Step Solution

This problem requires us to find the point of intersection of a given line and a plane, and then calculate the square of the distance from this intersection point to another specified point.

Step 1: Express the Line in Parametric Form The given equation of the line is in symmetric form: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}$ To find any point on this line, we set this expression equal to a parameter $\lambda$. This allows us to represent the coordinates of any point on the line in terms of a single variable $\lambda$. $\frac{x - 1}{2} = \lambda \implies x = 2\lambda + 1$ $\frac{y - 2}{3} = \lambda \implies y = 3\lambda + 2$ $\frac{z + 1}{6} = \lambda \implies z = 6\lambda - 1$ Thus, any point on the line can be represented as $P(2\lambda + 1, 3\lambda + 2, 6\lambda - 1)$.

Step 2: Find the Point of Intersection of the Line and the Plane The equation of the plane is given as: $2x - y + z = 6$ For a point to be the intersection of the line and the plane, it must satisfy both equations. We substitute the parametric coordinates of a general point on the line (from Step 1) into the plane's equation. This will give us a linear equation in $\lambda$, which we can solve to find the specific value of $\lambda$ corresponding to the intersection point. Substitute $x = 2\lambda + 1$, $y = 3\lambda + 2$, and $z = 6\lambda - 1$ into the plane equation: $2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$ Now, we simplify and solve for $\lambda$: $4\lambda + 2 - 3\lambda - 2 + 6\lambda - 1 = 6$ Combine the $\lambda$ terms and the constant terms: $(4\lambda - 3\lambda + 6\lambda) + (2 - 2 - 1) = 6$ $7\lambda - 1 = 6$ $7\lambda = 7$ $\lambda = 1$ Now that we have the value of $\lambda$, we substitute it back into the parametric equations of the line to find the exact coordinates of the point of intersection, $P_I$. For $\lambda = 1$: $x = 2(1) + 1 = 3$ $y = 3(1) + 2 = 5$ $z = 6(1) - 1 = 5$ So, the point of intersection of the line and the plane is $P_I(3, 5, 5)$.

Step 3: Calculate the Square of the Distance We need to find the square of the distance between the intersection point $P_I(3, 5, 5)$ and the given point $P_G(-1, -1, 2)$. Using the distance formula for the square of the distance $d^2$: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$ Let $(x_1, y_1, z_1) = (3, 5, 5)$ and $(x_2, y_2, z_2) = (-1, -1, 2)$. $d^2 = ((-1) - 3)^2 + ((-1) - 5)^2 + (2 - 5)^2$ $d^2 = (-4)^2 + (-6)^2 + (-3)^2$ $d^2 = 16 + 36 + 9$ $d^2 = 61$ The square of the distance is 61.

Common Mistakes & Tips

• Parametric Form: Always convert the line equation to its parametric form. This is the most reliable method for finding points on a line and for intersection problems.
• Algebraic Accuracy: Pay close attention to signs and arithmetic during the substitution and simplification steps. A small calculation error can lead to an incorrect value of $\lambda$ and consequently an incorrect intersection point.
• Read the Question Carefully: Note that the question asks for the square of the distance, not the distance itself. This means the final step does not require taking a square root.

Summary

This problem is a classic example of finding the intersection of a line and a plane in 3D geometry and then calculating the distance to another point. The process involves converting the line to its parametric form, substituting these parametric equations into the plane equation to find the parameter $\lambda$, and then using $\lambda$ to determine the exact coordinates of the intersection point. Finally, the 3D distance formula is applied to find the square of the distance between the intersection point and the given point. Following these steps systematically ensures an accurate solution.

The final answer is $\boxed{61}$.