1. Key Concepts and Formulas

• Equation of a Line in Parametric Form: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $a,b,c$ can be represented as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$. Any general point on this line can then be expressed as $(x_0+a\lambda, y_0+b\lambda, z_0+c\lambda)$. This form is crucial for identifying points on the line and for finding intersections.

• Intersection of Two Lines: To find the point of intersection of two lines, we express each line in its parametric form using different parameters (e.g., $s$ and $t$). By equating the corresponding coordinates ($x, y, z$) of the general points on each line, we form a system of linear equations. Solving this system yields the values of the parameters, which are then substituted back into either parametric equation to find the coordinates of the unique intersection point.

• Shortest Distance from a Point to a Line: The shortest distance from a point $P(x_1, y_1, z_1)$ to a line $\vec{r} = \vec{a} + \lambda \vec{d}$ is the perpendicular distance. This distance is found by:

  1. Representing a general point $R$ on the line in terms of its parameter (e.g., $\lambda$).
  2. Forming the vector $\vec{PR}$ connecting the given point $P$ to the general point $R$.
  3. Applying the condition that $\vec{PR}$ must be perpendicular to the direction vector $\vec{d}$ of the line. This means their dot product is zero: $\vec{PR} \cdot \vec{d} = 0$.
  4. Solving this equation for $\lambda$ to find the coordinates of the specific point $R$ (the foot of the perpendicular) on the line that is closest to $P$.
  5. Calculating the distance between $P$ and $R$ using the 3D distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

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2. Step-by-Step Solution

Step 1: Determine the Point of Intersection, P.

• Purpose: Our first goal is to find the coordinates of point P, which is the intersection of the two given lines.
• Method: We will express each line in its parametric form using distinct parameters, equate their coordinates, and solve the resulting system of equations.

Let the two given lines be $L_1$ and $L_2$: Line $L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ Line $L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$

For Line $L_1$, let $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}=s$. A general point on $L_1$ can be written as $(x, y, z) = (2+s, 4+5s, 2+s)$.

For Line $L_2$, let $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}=t$. A general point on $L_2$ can be written as $(x, y, z) = (3+2t, 2+3t, 3+2t)$.

At the point of intersection $P$, the coordinates must be the same for both lines. Therefore, we equate the corresponding coordinates:

1. $2+s = 3+2t \quad \implies s - 2t = 1 \quad \text{(Equation 1)}$
2. $4+5s = 2+3t \quad \implies 5s - 3t = -2 \quad \text{(Equation 2)}$
3. $2+s = 3+2t \quad \implies s - 2t = 1 \quad \text{(Equation 3)}$ Notice that Equation 1 and Equation 3 are identical, indicating a consistent system for two intersecting lines. We solve the system formed by Equation 1 and Equation 2.

From Equation 1, we can express $s$ in terms of $t$: $s = 1+2t$. Substitute this expression for $s$ into Equation 2: $5(1+2t) - 3t = -2$ $5 + 10t - 3t = -2$ $7t = -7$ $t = -1$

Now, substitute the value of $t$ back into the expression for $s$: $s = 1+2(-1) = 1-2 = -1$

We have found the parameter values $s=-1$ and $t=-1$. We can find the coordinates of point $P$ by substituting either $s$ into the parametric form of $L_1$ or $t$ into the parametric form of $L_2$. Using $s=-1$ for $L_1$: $x_P = 2+(-1) = 1$ $y_P = 4+5(-1) = 4-5 = -1$ $z_P = 2+(-1) = 1$ Thus, the point of intersection $P$ is $(1, -1, 1)$.

(Self-check): Using $t=-1$ for $L_2$: $x_P = 3+2(-1)=1$, $y_P = 2+3(-1)=-1$, $z_P = 3+2(-1)=1$. The coordinates match, confirming $P(1, -1, 1)$.

Step 2: Express the Target Line in Symmetric Form and Identify its Direction Vector.

• Purpose: To calculate the shortest distance, we need the target line to be in a standard form that clearly shows a point on the line and its direction vector.
• Method: We convert the given continuous equality $4x=2y=z$ into the symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$.

The given line is $4x=2y=z$. To convert it to symmetric form, we can divide by the least common multiple of the coefficients of $x, y, z$ (which are 4, 2, 1, respectively), which is 4. $\frac{4x}{4} = \frac{2y}{4} = \frac{z}{4}$ $\frac{x}{1} = \frac{y}{2} = \frac{z}{4}$ This is the standard symmetric form of the line. From this, we can identify:

• A point on the line: $(0,0,0)$ (since $x_0=0, y_0=0, z_0=0$).
• The direction vector of the line: $\vec{d} = \langle 1, 2, 4 \rangle$.

Step 3: Find the Foot of the Perpendicular from P to the Target Line.

• Purpose: To find the shortest distance, we need to find the specific point $R$ on the target line that is closest to $P$. This point $R$ is called the foot of the perpendicular.
• Method: We represent a general point $R$ on the target line using a parameter, form the vector $\vec{PR}$, and use the perpendicularity condition ($\vec{PR} \cdot \vec{d} = 0$) to solve for the parameter and find $R$.

Let $P$ be $(1, -1, 1)$. A general point $R$ on the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{4} = \lambda$ can be written as $R(\lambda, 2\lambda, 4\lambda)$.

Now, we form the vector $\vec{PR}$: $\vec{PR} = R - P = \langle \lambda - 1, 2\lambda - (-1), 4\lambda - 1 \rangle = \langle \lambda - 1, 2\lambda + 1, 4\lambda - 1 \rangle$.

The direction vector of the line is $\vec{d} = \langle 1, 2, 4 \rangle$. For $\vec{PR}$ to be the shortest distance, it must be perpendicular to the line, and thus perpendicular to its direction vector $\vec{d}$. So, their dot product must be zero: $\vec{PR} \cdot \vec{d} = 0$. $(\lambda - 1)(1) + (2\lambda + 1)(2) + (4\lambda - 1)(4) = 0$ $\lambda - 1 + 4\lambda + 2 + 16\lambda - 4 = 0$ Combine the terms: $(1+4+16)\lambda + (-1+2-4) = 0$ $21\lambda - 3 = 0$ $21\lambda = 3$ $\lambda = \frac{3}{21} = \frac{1}{7}$

Now, substitute this value of $\lambda$ back into the coordinates of $R$ to find the foot of the perpendicular: $R_x = \frac{1}{7}$ $R_y = 2\left(\frac{1}{7}\right) = \frac{2}{7}$ $R_z = 4\left(\frac{1}{7}\right) = \frac{4}{7}$ So, the foot of the perpendicular is $R\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right)$.

Step 4: Calculate the Shortest Distance PR.

• Purpose: The shortest distance is simply the distance between the point $P$ and the foot of the perpendicular $R$.
• Method: We use the 3D distance formula to calculate the distance between $P(1, -1, 1)$ and $R\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right)$.

The distance formula is $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. $PR = \sqrt{\left(\frac{1}{7} - 1\right)^2 + \left(\frac{2}{7} - (-1)\right)^2 + \left(\frac{4}{7} - 1\right)^2}$ $PR = \sqrt{\left(\frac{1-7}{7}\right)^2 + \left(\frac{2+7}{7}\right)^2 + \left(\frac{4-7}{7}\right)^2}$ $PR = \sqrt{\left(\frac{-6}{7}\right)^2 + \left(\frac{9}{7}\right)^2 + \left(\frac{-3}{7}\right)^2}$ $PR = \sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}}$ $PR = \sqrt{\frac{36+81+9}{49}}$ $PR = \sqrt{\frac{126}{49}}$

To simplify the radical, we can factorize $126 = 9 \times 14$: $PR = \sqrt{\frac{9 \times 14}{49}}$ $PR = \frac{\sqrt{9} \times \sqrt{14}}{\sqrt{49}}$ $PR = \frac{3\sqrt{14}}{7}$

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3. Common Mistakes & Tips

• Different Parameters: When finding the intersection of two lines, always use different parameters (e.g., $s$ and $t$) for each line to avoid algebraic confusion.
• Converting Line Equations: Be careful when converting line equations like $4x=2y=z$ into symmetric form. Ensure the coefficients of $x, y, z$ in the numerator are 1. The denominators then represent the direction ratios.
• Perpendicularity Condition: Remember that the shortest distance from a point to a line is along the perpendicular. This directly translates to the dot product of the connecting vector ($\vec{PR}$) and the line's direction vector ($\vec{d}$) being zero.
• Algebraic Precision: 3D geometry problems often involve fractions and square roots. Maintain careful algebraic manipulation throughout to avoid errors, especially during the final simplification of the radical.

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4. Summary

This problem required a two-stage approach. First, we found the point of intersection $P(1, -1, 1)$ of the two given lines by expressing them in parametric form and solving the system of equations. Second, we calculated the shortest distance from this point $P$ to the third line $4x=2y=z$. This was achieved by converting the third line to its symmetric form, identifying its direction vector, finding the foot of the perpendicular $R\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right)$ using the dot product condition, and finally computing the distance between $P$ and $R$. The shortest distance was found to be $\frac{3\sqrt{14}}{7}$.

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5. Final Answer

The final answer is $\boxed{\frac{3 \sqrt{14}}{7}}$, which corresponds to option (A).