Key Concepts and Formulas

1. Parametric Form of a Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be expressed in symmetric form as $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$. By setting this equal to a parameter (e.g., $\lambda$), any point on the line can be represented parametrically as $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$. This form is crucial for finding general points on a line.

2. Intersection of Two Lines: If two lines intersect, there is a common point that satisfies the equations of both lines. To find this point, we represent a general point on each line using different parameters. By equating the corresponding $x, y, z$ coordinates, we form a system of equations. Solving for the parameters and substituting them back into either line's parametric equations yields the intersection point. A crucial step is to verify the parameters satisfy all three coordinate equations to confirm an intersection.

3. Midpoint Formula: The midpoint $M$ of a line segment connecting two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the average of their respective coordinates: $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$

4. Distance of a Point from a Plane: The perpendicular distance $D$ of a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D'=0$ is calculated using the formula: $D = \frac{|Ax_0+By_0+Cz_0+D'|}{\sqrt{A^2+B^2+C^2}}$ Note that the constant term in the plane equation must be moved to the left side to match the $D'$ in the formula.

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Step-by-Step Solution

Step 1: Convert All Line Equations to Standard Parametric Form

The first step is to rewrite all given line equations in the standard symmetric form, $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$, and then convert them into their parametric forms. This makes it easy to represent any point on the line.

• Line 1 (The main line): $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ To get the standard form, we must rewrite the $y$-term as $(y-y_1)$: $\frac{x}{1}=\frac{-(y-6)}{2}=\frac{z+8}{5}$ This simplifies to $\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}$. Let's set this equal to a parameter $\lambda$. Any point on Line 1 can be written as: $P_1(\lambda) = (\lambda, 6-2\lambda, 5\lambda-8)$

• Line 2: $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ This line is already in the standard symmetric form. Let's set it equal to a parameter $k$. Any point on Line 2 can be written as: $P_2(k) = (4k+5, 3k+7, k-2)$

• Line 3: $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ Similar to Line 1, we rewrite the $y$-term: $\frac{x+3}{6}=\frac{-(y-3)}{3}=\frac{z-6}{1}$ This becomes $\frac{x+3}{6}=\frac{y-3}{-3}=\frac{z-6}{1}$. Let's set this equal to a parameter $\mu$. Any point on Line 3 can be written as: $P_3(\mu) = (6\mu-3, 3-3\mu, \mu+6)$ Explanation: We use distinct parameters ($\lambda, k, \mu$) for each line because they are independent.

Step 2: Find the Coordinates of Point A

Point A is the intersection of Line 1 and Line 2. This means the coordinates $P_1(\lambda)$ and $P_2(k)$ must be identical at point A. We equate their corresponding $x, y, z$ coordinates:

1. $x$-coordinate: $\lambda = 4k+5$
2. $y$-coordinate: $6-2\lambda = 3k+7$
3. $z$-coordinate: $5\lambda-8 = k-2$

We now solve this system of three equations for the two unknowns, $\lambda$ and $k$. Substitute equation (1) into equation (2): $6-2(4k+5) = 3k+7$ $6-8k-10 = 3k+7$ $-4-8k = 3k+7$ $-11k = 11 \Rightarrow k = -1$

Now, substitute $k=-1$ back into equation (1) to find $\lambda$: $\lambda = 4(-1)+5 = -4+5 = 1$

To confirm consistency, we substitute $\lambda=1$ and $k=-1$ into the third equation (3): $5(1)-8 = (-1)-2$ $5-8 = -3$ $-3 = -3$ The values are consistent, confirming that the lines intersect at a unique point A.

Now, substitute $\lambda=1$ into the parametric equations for Line 1 (or $k=-1$ into Line 2) to find the coordinates of A: $x_A = 1$ $y_A = 6-2(1) = 4$ $z_A = 5(1)-8 = -3$ Thus, the coordinates of point A are $(1, 4, -3)$.

Step 3: Find the Coordinates of Point B

Point B is the intersection of Line 1 and Line 3. We equate the coordinates $P_1(\lambda)$ and $P_3(\mu)$:

1. $x$-coordinate: $\lambda = 6\mu-3$
2. $y$-coordinate: $6-2\lambda = 3-3\mu$
3. $z$-coordinate: $5\lambda-8 = \mu+6$

Substitute equation (1) into equation (2): $6-2(6\mu-3) = 3-3\mu$ $6-12\mu+6 = 3-3\mu$ $12-12\mu = 3-3\mu$ $9 = 9\mu \Rightarrow \mu = 1$

Now, substitute $\mu=1$ back into equation (1) to find $\lambda$: $\lambda = 6(1)-3 = 3$

To confirm consistency, we substitute $\lambda=3$ and $\mu=1$ into the third equation (3): $5(3)-8 = (1)+6$ $15-8 = 7$ $7 = 7$ The values are consistent, confirming the intersection at point B.

Now, substitute $\lambda=3$ into the parametric equations for Line 1 (or $\mu=1$ into Line 3) to find the coordinates of B: $x_B = 3$ $y_B = 6-2(3) = 0$ $z_B = 5(3)-8 = 7$ Thus, the coordinates of point B are $(3, 0, 7)$.

Step 4: Find the Midpoint of Line Segment AB

Let M be the midpoint of the line segment AB. Using the midpoint formula with A$(1, 4, -3)$ and B$(3, 0, 7)$: $M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}, \frac{z_A+z_B}{2}\right)$ $M = \left(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}\right)$ $M = \left(\frac{4}{2}, \frac{4}{2}, \frac{4}{2}\right)$ $M = (2, 2, 2)$

Step 5: Calculate the Distance of Midpoint M from the Plane

The given plane equation is $2x-2y+z=14$. To use the distance formula, we must write it in the standard form $Ax+By+Cz+D'=0$: $2x-2y+z-14=0$. Here, the coefficients are $A=2, B=-2, C=1$, and the constant term is $D'=-14$. The midpoint M is $(x_0, y_0, z_0) = (2, 2, 2)$.

Using the distance formula: $D = \frac{|Ax_0+By_0+Cz_0+D'|}{\sqrt{A^2+B^2+C^2}}$ $D = \frac{|2(2) + (-2)(2) + 1(2) + (-14)|}{\sqrt{(2)^2 + (-2)^2 + (1)^2}}$ $D = \frac{|4 - 4 + 2 - 14|}{\sqrt{4 + 4 + 1}}$ $D = \frac{|-12|}{\sqrt{9}}$ $D = \frac{12}{3}$ $D = 4$

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Common Mistakes & Tips

• Standard Form Conversion: Always ensure line equations like $\frac{c-y}{b}$ are correctly converted to $\frac{y-c}{-b}$ to avoid sign errors in direction ratios.
• Parameter Verification: When finding line intersections, after solving for two parameters using two equations, always substitute them into the third equation to verify consistency. If they don't satisfy the third equation, the lines are skew and do not intersect.
• Plane Equation Form: For the distance formula, ensure the plane equation is in the form $Ax+By+Cz+D'=0$. If it's $Ax+By+Cz=D''$, rewrite it as $Ax+By+Cz-D''=0$.
• Absolute Value: Do not forget the absolute value in the numerator of the distance formula, as distance is always non-negative.

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Summary

This problem required a systematic application of 3D geometry concepts. We first converted all line equations into their parametric forms to easily represent general points. Next, we found the intersection points A and B by equating coordinates of the relevant lines and solving the resulting systems of linear equations, ensuring to verify consistency. After obtaining A and B, we calculated their midpoint M using the midpoint formula. Finally, we used the formula for the distance of a point from a plane to find the distance of M from the given plane, arriving at a distance of 4 units.

The final answer is $\boxed{4}$ which corresponds to option (C).