Key Concepts and Formulas

1. Equation of a Plane and Line Relationship:

   • A plane passing through the origin is given by $lx + my + nz = 0$, where $\langle l, m, n \rangle$ is its normal vector.
   • If a line $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ lies completely within a plane $Ax + By + Cz + D = 0$, then:
     • Any point $(x_1, y_1, z_1)$ on the line must satisfy the plane's equation.
     • The direction vector of the line $\langle a, b, c \rangle$ must be perpendicular to the normal vector of the plane $\langle A, B, C \rangle$. This implies their dot product is zero: $aA + bB + cC = 0$.

2. Section Formula (Internal Division): The coordinates of a point R that divides the line segment joining points A($x_1, y_1, z_1$) and B($x_2, y_2, z_2$) internally in the ratio $k:1$ are given by: $R = \left( \frac{kx_2 + 1x_1}{k+1}, \frac{ky_2 + 1y_1}{k+1}, \frac{kz_2 + 1z_1}{k+1} \right)$

3. Condition for a Point to Lie on a Plane: If a point $(x_0, y_0, z_0)$ lies on a plane $Ax + By + Cz + D = 0$, then its coordinates must satisfy the plane's equation: $Ax_0 + By_0 + Cz_0 + D = 0$.

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Step-by-Step Solution

Step 1: Extract Information from the Given Line and Plane

First, we identify the key features of the given plane and line. The plane P is given by $lx + my + nz = 0$. Its normal vector is $\vec{N} = \langle l, m, n \rangle$. Since the constant term is zero, this plane passes through the origin.

The given line is ${{1 - x} \over 1} = {{y + 4} \over 2} = {{z + 2} \over 3}$. To work with this line effectively, we must convert it into the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The term ${{1 - x} \over 1}$ needs to be rewritten as ${{-(x - 1)} \over 1} = {{x - 1} \over {-1}}$. The other terms are already in the correct form: ${{y - (-4)} \over 2}$ and ${{z - (-2)} \over 3}$. So, the line equation in standard form is: ${{x - 1} \over {-1}} = {{y - (-4)} \over 2} = {{z - (-2)} \over 3}$ From this, we can identify:

• A point on the line: $A_L = (1, -4, -2)$.
• The direction vector of the line: $\vec{d} = \langle -1, 2, 3 \rangle$.

Step 2: Apply Conditions for the Line to Lie in the Plane

Since the line lies completely within the plane P, two crucial conditions must be satisfied:

Condition 2a: The point $A_L$ on the line must lie on the plane P. Substitute the coordinates of $A_L(1, -4, -2)$ into the plane equation $lx + my + nz = 0$: $l(1) + m(-4) + n(-2) = 0$ $l - 4m - 2n = 0 \quad \text{ (Equation 1)}$

Condition 2b: The direction vector of the line $\vec{d}$ must be perpendicular to the normal vector of the plane $\vec{N}$. This means their dot product must be zero: $\vec{d} \cdot \vec{N} = 0$. $\langle -1, 2, 3 \rangle \cdot \langle l, m, n \rangle = 0$ $(-1)l + (2)m + (3)n = 0$ $-l + 2m + 3n = 0 \quad \text{ (Equation 2)}$

Step 3: Solve for the Ratios of $l, m, n$

We now have a system of two linear equations with three variables $l, m, n$:

1. $l - 4m - 2n = 0$
2. $-l + 2m + 3n = 0$

To find the ratios of $l, m, n$, we can solve this system. A common method is elimination: Add Equation 1 and Equation 2: $(l - 4m - 2n) + (-l + 2m + 3n) = 0$ $(l - l) + (-4m + 2m) + (-2n + 3n) = 0$ $-2m + n = 0 \implies n = 2m$

Now, substitute $n = 2m$ into Equation 1: $l - 4m - 2(2m) = 0$ $l - 4m - 4m = 0$ $l - 8m = 0 \implies l = 8m$

We have found the relationships between $l, m,$ and $n$: $l = 8m$ and $n = 2m$. This implies the ratio $l:m:n$ is $8m:m:2m$. Assuming $m \ne 0$ (if $m=0$, then $l=0$ and $n=0$, which would mean the normal vector is $\langle 0,0,0 \rangle$, which is not possible for a plane), we can divide by $m$: $l:m:n = 8:1:2$ Therefore, the normal vector of the plane P can be taken as $\vec{N} = \langle 8, 1, 2 \rangle$.

Step 4: Form the Equation of Plane P

Using the normal vector $\langle 8, 1, 2 \rangle$ and the general form $lx + my + nz = 0$ for a plane passing through the origin, the equation of plane P is: $8x + 1y + 2z = 0$ $8x + y + 2z = 0$

Step 5: Apply the Section Formula to find Point R

Let R be the point that divides the line segment AB joining points A($-3, -6, 1$) and B($2, 4, -3$) in the ratio $k:1$. Using the section formula for internal division: $R = \left( \frac{k(x_B) + 1(x_A)}{k+1}, \frac{k(y_B) + 1(y_A)}{k+1}, \frac{k(z_B) + 1(z_A)}{k+1} \right)$ Substituting the coordinates: $R = \left( \frac{k(2) + 1(-3)}{k+1}, \frac{k(4) + 1(-6)}{k+1}, \frac{k(-3) + 1(1)}{k+1} \right)$ $R = \left( \frac{2k - 3}{k+1}, \frac{4k - 6}{k+1}, \frac{-3k + 1}{k+1} \right)$

Step 6: Apply the Condition for Point R to Lie on Plane P and Solve for k

Since point R lies on the plane $8x + y + 2z = 0$, its coordinates must satisfy the plane's equation. Substitute the coordinates of R into the plane equation: $8\left( \frac{2k - 3}{k+1} \right) + \left( \frac{4k - 6}{k+1} \right) + 2\left( \frac{-3k + 1}{k+1} \right) = 0$ To eliminate the denominators, we multiply the entire equation by $(k+1)$. Note that $k \ne -1$ for the ratio to be meaningful. $8(2k - 3) + (4k - 6) + 2(-3k + 1) = 0$ Expand and simplify the terms: $(16k - 24) + (4k - 6) + (-6k + 2) = 0$ Combine the terms containing $k$: $16k + 4k - 6k = (20 - 6)k = 14k$ Combine the constant terms: $-24 - 6 + 2 = -30 + 2 = -28$ The equation simplifies to: $14k - 28 = 0$ Solve for $k$: $14k = 28$ $k = \frac{28}{14}$ $k = 2$

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Common Mistakes & Tips

• Standard Form of Line Equation: Always ensure the line equation is in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Pay close attention to signs; for example, $1-x$ must be rewritten as $-(x-1)$, changing the sign of the denominator.
• Section Formula with Negative Coordinates: Be very careful with signs when substituting negative coordinates into the section formula to avoid arithmetic errors.
• Ratio vs. Absolute Values: When finding the normal vector components $l, m, n$, remember you are finding their ratios, not their absolute values. Any non-zero multiple of $\langle 8, 1, 2 \rangle$ would represent the same normal direction and thus the same plane.

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Summary

This problem required a systematic application of multiple 3D geometry concepts. We began by converting the given line equation to its standard form to easily identify a point on the line and its direction vector. Then, using the conditions for a line to lie in a plane (a point on the line must satisfy the plane equation, and the line's direction vector must be perpendicular to the plane's normal vector), we formed a system of equations to determine the ratios of the normal vector components $(l:m:n)$ for plane P. This allowed us to write the equation of plane P. Finally, we used the section formula to find the coordinates of the point R that divides the line segment AB in the ratio $k:1$. By substituting these coordinates into the plane P's equation, we solved for the value of $k$. The calculated value of $k$ is 2.

The final answer is $\boxed{2}$, which corresponds to option (A).