Key Concepts and Formulas

1. Parametric Form of a Line in 3D: A line given in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ can be expressed parametrically. By setting each ratio equal to a parameter (say, $k$), any point $(x,y,z)$ on the line can be represented as $x = x_0 + ak$, $y = y_0 + bk$, $z = z_0 + ck$. This form is essential for finding specific points on the line or for checking if a point lies on the line.
2. Point of Intersection of Two Lines in 3D: To find the intersection point of two lines, we represent each line using its parametric form but with different parameters (e.g., $k$ for the first line and $\alpha$ for the second). If the lines intersect, there must be a unique point common to both. This means that for specific values of $k$ and $\alpha$, the $x, y, z$ coordinates from both parametric forms must be equal. Equating these coordinates leads to a system of three linear equations in two variables ($k$ and $\alpha$). Solving any two equations for $k$ and $\alpha$ and then verifying these values with the third equation confirms if the lines intersect and provides the parameter values for the intersection point.
3. Distance Formula in 3D: The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

---

Step-by-Step Solution

Step 1: Expressing the Lines in Parametric Form Our first task is to convert the given symmetric equations of the lines into their parametric forms. This allows us to represent any general point on each line using a single variable.

For the first line: $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ Let's set each ratio equal to a parameter, say $k$: $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1} = k$ From this, we can write the coordinates of any point on the first line in terms of $k$: $x+6 = 3k \implies x = 3k-6$ $y = 2k$ $z+1 = k \implies z = k-1$ So, any point on the first line can be denoted as $P_1(k) = (3k-6, 2k, k-1)$.

For the second line: $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ We introduce a different parameter, say $\alpha$, for the second line to avoid confusion: $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2} = \alpha$ From this, we can write the coordinates of any point on the second line in terms of $\alpha$: $x-7 = 4\alpha \implies x = 4\alpha+7$ $y-9 = 3\alpha \implies y = 3\alpha+9$ $z-4 = 2\alpha \implies z = 2\alpha+4$ So, any point on the second line can be denoted as $P_2(\alpha) = (4\alpha+7, 3\alpha+9, 2\alpha+4)$.

Step 2: Setting up Equations for Intersection If the two lines intersect, there must be a specific point $(x,y,z)$ that lies on both lines. This means that for some particular values of $k$ and $\alpha$, the coordinates of $P_1(k)$ and $P_2(\alpha)$ must be identical. We equate the corresponding $x, y, z$ coordinates:

1. Equating x-coordinates: $3k-6 = 4\alpha+7$
2. Equating y-coordinates: $2k = 3\alpha+9$
3. Equating z-coordinates: $k-1 = 2\alpha+4$

Step 3: Solving the System of Equations Now we rearrange the three equations to form a standard system of linear equations in terms of $k$ and $\alpha$: (1) $3k - 4\alpha = 13$ (2) $2k - 3\alpha = 9$ (3) $k - 2\alpha = 5$

We can solve any two of these equations simultaneously to find the values of $k$ and $\alpha$. Let's use equations (1) and (2) using the elimination method. Multiply equation (1) by 2 and equation (2) by 3 to make the coefficients of $k$ equal: $2 \times (3k - 4\alpha = 13) \implies 6k - 8\alpha = 26$ $3 \times (2k - 3\alpha = 9) \implies 6k - 9\alpha = 27$

Subtract the second modified equation from the first modified equation: $(6k - 8\alpha) - (6k - 9\alpha) = 26 - 27$ $6k - 8\alpha - 6k + 9\alpha = -1$ $\alpha = -1$

Now substitute the value of $\alpha = -1$ into equation (2) to find $k$: $2k - 3(-1) = 9$ $2k + 3 = 9$ $2k = 6$ $k = 3$

Verification (Crucial Step): It is vital to verify these values of $k$ and $\alpha$ with the third equation. If they satisfy the third equation, the lines intersect at a unique point. If not, the lines are skew (non-parallel and non-intersecting). Using equation (3): $k - 2\alpha = 5$ Substitute $k=3$ and $\alpha=-1$: $3 - 2(-1) = 3 + 2 = 5$ Since $5=5$, the values $k=3$ and $\alpha=-1$ are consistent with all three equations, confirming that the lines intersect.

Step 4: Determining the Intersection Point Now that we have the values of $k$ and $\alpha$, we can find the coordinates of the intersection point. We can substitute $k=3$ into the parametric form of the first line, $P_1(k)$, or substitute $\alpha=-1$ into the parametric form of the second line, $P_2(\alpha)$. Both methods must yield the same point.

Using $P_1(k)$ with $k=3$: $x = 3(3)-6 = 9-6 = 3$ $y = 2(3) = 6$ $z = 3-1 = 2$ The point of intersection is $(3,6,2)$.

(As a quick check using $P_2(\alpha)$ with $\alpha=-1$: $x = 4(-1)+7 = -4+7 = 3$ $y = 3(-1)+9 = -3+9 = 6$ $z = 2(-1)+4 = -2+4 = 2$ This confirms the intersection point is $(3,6,2)$.)

Step 5: Calculating the Distance We need to find the distance $d$ of the intersection point $(3,6,2)$ from the given point $(7,8,9)$. Let $(x_1, y_1, z_1) = (3,6,2)$ and $(x_2, y_2, z_2) = (7,8,9)$. Using the 3D distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ $d = \sqrt{(7-3)^2 + (8-6)^2 + (9-2)^2}$ $d = \sqrt{(4)^2 + (2)^2 + (7)^2}$ $d = \sqrt{16 + 4 + 49}$ $d = \sqrt{69}$

Step 6: Final Calculation The problem asks for the value of $d^2+6$. We found $d = \sqrt{69}$, so $d^2 = 69$. $d^2+6 = 69+6 = 75$

---

Common Mistakes & Tips

• Using the Same Parameter: A common mistake is using the same parameter (e.g., $k$) for both lines when finding their intersection. This is incorrect because the parameter values for the intersection point will generally be different for each line. Always use distinct parameters (e.g., $k$ and $\alpha$).
• Forgetting to Verify: After solving for the parameters using two of the three equations, it is crucial to substitute these values into the third equation. If the third equation is not satisfied, it means the lines are skew and do not intersect. In such a scenario, there would be no point of intersection.
• Arithmetic Errors: Be careful with calculations while solving the system of equations and applying the distance formula. Simple sign errors or calculation mistakes are frequent pitfalls.
• Tip for Checking Intersection Point: Once you find the intersection point, substitute its coordinates back into the original symmetric equations of both lines. If the point lies on both lines, it will satisfy both equations.

---

Summary

This problem involved a standard sequence of steps in 3D geometry: first, converting the given lines from symmetric to parametric form using distinct parameters. Next, we found the point of intersection by equating the corresponding coordinates of the parametric forms, which led to a system of linear equations. We solved this system for the parameters and crucially verified the solution with the third equation to confirm intersection. Substituting the parameter back into either line's parametric form gave us the coordinates of the intersection point. Finally, we applied the 3D distance formula between this intersection point and the given point, and then evaluated the required expression $d^2+6$.

The final answer is $\boxed{75}$, which corresponds to option (A).