If the distances of the point (1, 2, a) from the line (x-1)/1 = y/2 = (z-1)/1 along the lines L1 : (

Q: Wait, why are we setting the coordinates of $A$ equal to the line $L$?

Great question! Think of it this way: point $A$ is the meeting point. It is defined by the path $L_1$, but it is also a point on the line $L$. Therefore, its coordinates must satisfy the equation of $L$. By equating the expressions, we are essentially forcing the path to intersect the target line.

Q: Can we just use the perpendicular distance formula for this?

No, that is a classic trap! The problem specifically asks for the distance *along* the lines $L_1$ and $L_2$. This is not the shortest (perpendicular) distance. It is the length of the segment connecting $P$ to the intersection point on $L$ along a specific direction. Always read the problem carefully!

Q: Why does the $a^2$ term cancel out in the final step?

It is a beautiful piece of algebraic symmetry! When you expand $(4-a)^2$ and $a^2$, the $a^2$ terms appear on both sides of the equation. This is a sign that the geometry is consistent and that the problem is well-posed. It simplifies the quadratic equation into a linear one, making your life much easier.

Question

If the distances of the point $(1, 2, a)$ from the line $\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$ along the lines $L_1 : \frac{x-1}{3} = \frac{y-2}{4} = \frac{z-a}{b}$ and $L_2 : \frac{x-1}{1} = \frac{y-2}{4} = \frac{z-a}{c}$ are equal, then $a + b + c$ is equal to

Options:

(A) 5

(B) 7

(C) 4

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You see a point $P(1, 2, a)$ suspended in the air, and below it, a line $L$ defined by the equation:

$$\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$$

We are tasked with finding the distance from $P$ to $L$ along two specific, predefined paths, $L_1$ and $L_2$, under the condition that these distances are equal.

### The First Encounter: Finding Point $A$

Let us focus on the first path, $L_1$, defined by:

$$\frac{x-1}{3} = \frac{y-2}{4} = \frac{z-a}{b} = \lambda$$

Any point on this line can be described by the parameter $\lambda$. Let the intersection point be $A$, with coordinates $(3\lambda + 1, 4\lambda + 2, b\lambda + a)$.

Because $A$ also lies on the line $L$, its coordinates must satisfy the equation of $L$. Substituting these coordinates into the equation of $L$, we get:

$$\frac{(3\lambda+1)-1}{1} = \frac{4\lambda+2}{2} = \frac{(b\lambda+a)-1}{1}$$

Simplifying this, we find:

$$\frac{3\lambda}{1} = \frac{4\lambda+2}{2} = \frac{b\lambda+a-1}{1}$$

By equating the first two terms, $3\lambda = 2\lambda + 1$, we find that $\lambda = 1$. Substituting $\lambda = 1$ into the third term, we discover the elegant relation $a + b = 4$. Our point $A$ is now fixed at $(4, 6, 4)$.

### The Second Encounter: Finding Point $B$

Now, we repeat this process for the second path, $L_2$, defined by:

$$\frac{x-1}{1} = \frac{y-2}{4} = \frac{z-a}{c} = \mu$$

Let the intersection point be $B$, with coordinates $(\mu + 1, 4\mu + 2, c\mu + a)$. Since $B$ lies on $L$, we substitute its coordinates into the equation of $L$:

$$\frac{(\mu+1)-1}{1} = \frac{4\mu+2}{2} = \frac{(c\mu+a)-1}{1}$$

This simplifies to:

$$\frac{\mu}{1} = \frac{4\mu+2}{2} = \frac{c\mu+a-1}{1}$$

Equating the first two terms, $\mu = 2\mu + 1$, we find $\mu = -1$. Substituting $\mu = -1$ into the third term, we get $-1 = -c + a - 1$, which simplifies to $a = c$. Our point $B$ is $(0, -2, 0)$.

### The Climax: The Distance Equality

The problem states that the distances from $P$ to $L$ along these two paths are equal, meaning $PA = PB$. To make our calculations cleaner, we work with the squares of the distances: $PA^2 = PB^2$.

Using the distance formula:

$$PA^2 = (4-1)^2 + (6-2)^2 + (4-a)^2 = 25 + (4-a)^2$$

$$PB^2 = (0-1)^2 + (-2-2)^2 + (0-a)^2 = 17 + a^2$$

Setting them equal, we have:

$$25 + 16 - 8a + a^2 = 17 + a^2$$

The $a^2$ terms cancel out, leaving us with $41 - 8a = 17$, or $8a = 24$. Thus, **$a = 3$**.

With $a = 3$, we find $b = 4 - 3 = 1$ and $c = 3$. The final sum is:

$$a + b + c = 3 + 1 + 3 = 7$$

We have navigated the 3D space, solved the intersections, and arrived at the elegant solution of **$7$**.

Select Answer:

Visualized Solution

Visualizing the Setup

Path 1: Line $L_{1}$

Intersection of $A$ and $L$

Solving for $λ$ and $b$

Path 2: Line $L_{2}$

Intersection of $B$ and $L$

Solving for $μ$ and $c$

The Distance Equality

Applying the Distance Formula

Solving for $a$

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The First Encounter

The Second Encounter

The Climax

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