Key Concepts and Formulas

1. Equation of a Line in 3D:

   • Two-Point Form: A line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ has the symmetric equation: $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$
   • Parametric Form: If the line is given as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$, then any point on the line can be represented as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$. The vector $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction vector of the line.

2. Foot of the Perpendicular from a Point to a Line: If N is the foot of the perpendicular from a point P to a line, then the vector $\vec{PN}$ is perpendicular to the direction vector of the line. This implies their dot product is zero: $\vec{PN} \cdot \vec{d} = 0$.

3. Distance of a Point from a Plane: The perpendicular distance of a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$ is given by the formula: $\text{Distance} = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$

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Step-by-Step Solution

Step 1: Find the Equation of the Line and its Parametric Form

We are given two points on the line: $A(4,5,8)$ and $B(1,-7,5)$. Our first goal is to express any point on this line using a parameter, which will allow us to represent the foot of the perpendicular, N.

• Why this step? To define the coordinates of the foot of the perpendicular N algebraically, we must first establish the algebraic representation of the line it lies on.

Using the two-point form of the line equation with $(x_1, y_1, z_1) = (4,5,8)$ and $(x_2, y_2, z_2) = (1,-7,5)$: $\frac{x-4}{1-4} = \frac{y-5}{-7-5} = \frac{z-8}{5-8}$ Simplifying the denominators, we get: $\frac{x-4}{-3} = \frac{y-5}{-12} = \frac{z-8}{-3}$ To work with simpler (positive) direction ratios, we can multiply the denominators by $-1$: $\frac{x-4}{3} = \frac{y-5}{12} = \frac{z-8}{3}$ Let this common ratio be $\lambda$. This is the parametric form of the line: $\frac{x-4}{3} = \frac{y-5}{12} = \frac{z-8}{3} = \lambda$ From this, any point N on the line can be expressed in terms of $\lambda$: $x_N = 3\lambda + 4$ $y_N = 12\lambda + 5$ $z_N = 3\lambda + 8$ So, the coordinates of the foot of the perpendicular N are $(3\lambda+4, 12\lambda+5, 3\lambda+8)$. The direction vector of the line, $\vec{d}$, is $3\hat{i} + 12\hat{j} + 3\hat{k}$.

Step 2: Form the Vector $\vec{PN}$

We are given the point $P(1,-2,3)$ from which the perpendicular is drawn. We need to form the vector connecting P to N.

• Why this step? The condition for N being the foot of the perpendicular involves the vector $\vec{PN}$. We need to express this vector in terms of $\lambda$ to solve for it.

The vector $\vec{PN}$ is calculated by subtracting the coordinates of P from the coordinates of N: $\vec{PN} = (x_N - x_P)\hat{i} + (y_N - y_P)\hat{j} + (z_N - z_P)\hat{k}$ Substituting the coordinates of P and N: $\vec{PN} = ((3\lambda+4) - 1)\hat{i} + ((12\lambda+5) - (-2))\hat{j} + ((3\lambda+8) - 3)\hat{k}$ $\vec{PN} = (3\lambda+3)\hat{i} + (12\lambda+7)\hat{j} + (3\lambda+5)\hat{k}$

Step 3: Apply the Perpendicularity Condition to Find $\lambda$

Since N is the foot of the perpendicular from P to the line, the vector $\vec{PN}$ must be perpendicular to the direction vector of the line, $\vec{d} = 3\hat{i} + 12\hat{j} + 3\hat{k}$.

• Why this step? This geometric condition (perpendicularity) is the key to finding the specific value of $\lambda$ that uniquely defines the point N.

The dot product of two perpendicular vectors is zero: $\vec{PN} \cdot \vec{d} = 0$. $(3\lambda+3)(3) + (12\lambda+7)(12) + (3\lambda+5)(3) = 0$ Expand and simplify the equation: $(9\lambda + 9) + (144\lambda + 84) + (9\lambda + 15) = 0$ Combine like terms: $(9+144+9)\lambda + (9+84+15) = 0$ $162\lambda + 108 = 0$ Solve for $\lambda$: $162\lambda = -108$ $\lambda = \frac{-108}{162}$ Dividing both numerator and denominator by their greatest common divisor (54): $\lambda = \frac{-2}{3}$

Step 4: Determine the Coordinates of N

Now that we have the value of $\lambda$, we can find the exact coordinates of the foot of the perpendicular, N.

• Why this step? The ultimate goal is to find the distance of N from a plane, so we must first determine N's precise location.

Substitute $\lambda = -\frac{2}{3}$ back into the parametric coordinates of N from Step 1: $x_N = 3\left(-\frac{2}{3}\right) + 4 = -2 + 4 = 2$ $y_N = 12\left(-\frac{2}{3}\right) + 5 = -8 + 5 = -3$ $z_N = 3\left(-\frac{2}{3}\right) + 8 = -2 + 8 = 6$ Thus, the coordinates of the foot of the perpendicular N are $(2,-3,6)$.

Step 5: Calculate the Distance of N from the Plane

Finally, we need to find the distance of the point $N(2,-3,6)$ from the given plane $2x-2y+z+5=0$.

• Why this step? This is the final requirement of the problem statement, applying the standard formula for point-to-plane distance.

Using the distance formula for a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$: $\text{Distance} = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ Here, $(x_0, y_0, z_0) = (2,-3,6)$ and the plane is $2x-2y+z+5=0$, so $A=2, B=-2, C=1, D=5$. Substitute these values into the formula: $\text{Distance} = \frac{|2(2) - 2(-3) + 1(6) + 5|}{\sqrt{2^2 + (-2)^2 + 1^2}}$ $\text{Distance} = \frac{|4 + 6 + 6 + 5|}{\sqrt{4 + 4 + 1}}$ $\text{Distance} = \frac{|21|}{\sqrt{9}}$ $\text{Distance} = \frac{21}{3}$ $\text{Distance} = 7$

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Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when substituting coordinates, especially negative ones, and during calculations like dot products. A single sign error can propagate and lead to an incorrect final answer.
• Direction Ratios: Ensure consistency when calculating direction ratios $(x_2-x_1, y_2-y_1, z_2-z_1)$. The order of subtraction must be maintained for all components.
• Absolute Value in Distance Formula: Always remember to take the absolute value of the numerator in the distance formula from a point to a plane, as distance must be non-negative.

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Summary

This problem is a comprehensive application of 3D coordinate geometry, requiring a sequential approach. We first established the parametric equation of the line passing through the two given points. Next, we expressed the foot of the perpendicular (N) in terms of a parameter $\lambda$. By utilizing the geometric condition that the vector $\vec{PN}$ is perpendicular to the line's direction vector (i.e., their dot product is zero), we solved for $\lambda$. This allowed us to find the exact coordinates of N. Finally, we applied the standard formula to calculate the perpendicular distance from point N to the given plane. This problem effectively tests understanding of line equations, vector perpendicularity, and point-to-plane distance.

The final answer is $\boxed{7}$, which corresponds to option (A).