Key Concepts and Formulas

1. Parametric Equation of a Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be represented in parametric form as $x = x_1 + at$, $y = y_1 + bt$, $z = z_1 + ct$, where $t$ is a parameter. Any point on the line can be expressed using this parameter.
2. Foot of the Perpendicular: If P is the foot of the perpendicular from an external point A to a line L, then the vector $\vec{AP}$ is perpendicular to the direction vector $\vec{d}$ of the line L. This implies their dot product is zero: $\vec{AP} \cdot \vec{d} = 0$.
3. Distance of a Point from a Plane: The perpendicular distance from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is given by the formula: $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Step-by-Step Solution

Step 1: Expressing a General Point on the Line in Parametric Form

The first step is to represent any arbitrary point on the given line using a single parameter. This is crucial for later algebraic manipulation to find the specific point P.

The given line L is in symmetric form: $L: \frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ To obtain the parametric form, we equate each ratio to a parameter, say $t$: $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3} = t$ From this, we can express the coordinates $(x, y, z)$ of any point on the line in terms of $t$:

• $x + 2 = 4t \implies x = 4t - 2$
• $y - 1 = 2t \implies y = 2t + 1$
• $z + 1 = 3t \implies z = 3t - 1$

Since P is the foot of the perpendicular from point A(1, 2, 4) onto line L, P must lie on line L. Therefore, the coordinates of P can be written as: $P = (4t - 2, 2t + 1, 3t - 1)$

Step 2: Applying the Perpendicularity Condition to Find the Parameter 't'

The defining characteristic of the foot of the perpendicular P is that the line segment AP is perpendicular to the line L. In vector terms, the vector $\vec{AP}$ must be perpendicular to the direction vector of line L. Their dot product must be zero.

Let the external point be $A = (1, 2, 4)$. The coordinates of point P are $(4t - 2, 2t + 1, 3t - 1)$.

First, we find the vector $\vec{AP}$: $\vec{AP} = (x_P - x_A)\hat{i} + (y_P - y_A)\hat{j} + (z_P - z_A)\hat{k}$ $\vec{AP} = ((4t - 2) - 1)\hat{i} + ((2t + 1) - 2)\hat{j} + ((3t - 1) - 4)\hat{k}$ $\vec{AP} = (4t - 3)\hat{i} + (2t - 1)\hat{j} + (3t - 5)\hat{k}$

Next, we identify the direction vector of the line L. From the symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, the direction vector is $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$. For the given line $L: \frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$, the direction vector is: $\vec{d} = 4\hat{i} + 2\hat{j} + 3\hat{k}$

Since $\vec{AP}$ is perpendicular to $\vec{d}$, their dot product is zero: $\vec{AP} \cdot \vec{d} = 0$ $( (4t - 3)\hat{i} + (2t - 1)\hat{j} + (3t - 5)\hat{k} ) \cdot ( 4\hat{i} + 2\hat{j} + 3\hat{k} ) = 0$ $4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$ Now, we solve this linear equation for $t$: $16t - 12 + 4t - 2 + 9t - 15 = 0$ $29t - 29 = 0$ $29t = 29$ $t = 1$

Step 3: Determining the Coordinates of the Foot of the Perpendicular, P

With the value of the parameter $t$ determined, we can now substitute it back into the parametric coordinates of P to find its exact location.

Recall the coordinates of P in terms of $t$: $P = (4t - 2, 2t + 1, 3t - 1)$ Substitute $t = 1$ into these expressions: $P = (4(1) - 2, 2(1) + 1, 3(1) - 1)$ $P = (4 - 2, 2 + 1, 3 - 1)$ $P = (2, 3, 2)$ So, the foot of the perpendicular P is $(2, 3, 2)$.

Step 4: Calculating the Distance from Point P to the Given Plane

The final step is to calculate the shortest distance from the point P, which we just found, to the given plane. We will use the standard formula for the distance from a point to a plane.

The point is $P = (x_0, y_0, z_0) = (2, 3, 2)$. The given plane equation is $3x + 4y + 12z + 23 = 0$. Here, the coefficients are $A = 3$, $B = 4$, $C = 12$, and the constant term is $D = 23$.

Using the distance formula: $D_{plane} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Substitute the values: $D_{plane} = \frac{|(3)(2) + (4)(3) + (12)(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}}$ Calculate the numerator: $|6 + 12 + 24 + 23| = |65| = 65$ Calculate the denominator: $\sqrt{9 + 16 + 144} = \sqrt{25 + 144} = \sqrt{169} = 13$ Now, divide the numerator by the denominator: $D_{plane} = \frac{65}{13}$ $D_{plane} = 5$

Thus, the distance of P from the plane $3x + 4y + 12z + 23 = 0$ is 5 units.

Common Mistakes & Tips

• Direction Vector Identification: Ensure you correctly extract the direction ratios from the line equation. For $\frac{x+a}{p}$, the direction ratio is $p$, not $-a$.
• Arithmetic Errors: Be meticulous with calculations, especially during vector subtraction, dot product, and substitution into the distance formula. Small errors can lead to incorrect final answers.
• Absolute Value in Distance Formula: Remember the absolute value in the numerator of the point-to-plane distance formula, as distance must always be non-negative.

Summary

This problem efficiently tests multiple core concepts in 3D geometry. We began by parametrizing the given line to represent the foot of the perpendicular, P. Then, using the perpendicularity condition ($\vec{AP} \cdot \vec{d} = 0$), we found the specific parameter value that defines P. Substituting this value yielded the coordinates of P. Finally, we applied the standard formula to calculate the distance from this point P to the given plane. This systematic approach is vital for solving such problems.

The final answer is $\boxed{5}$, which corresponds to option (A).