Key Concepts and Formulas

1. Parametric Form of a Line: A point on a line passing through $(x_0, y_0, z_0)$ with direction ratios $(a,b,c)$ can be represented as $(x_0 + \lambda a, y_0 + \lambda b, z_0 + \lambda c)$.
2. Direction Ratios (DRs) of a Line Segment: For two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, the direction ratios of the line segment $\vec{PQ}$ are $(x_2-x_1, y_2-y_1, z_2-z_1)$.
3. Perpendicular Distance from a Point to a Line: The length of the perpendicular $l$ from a point $A$ to a line passing through a point $P_0$ with direction vector $\vec{d}$ is given by $l = \frac{|\vec{P_0 A} \times \vec{d}|}{|\vec{d}|}$. Alternatively, if the vector $\vec{P_0 A}$ is perpendicular to $\vec{d}$, then $l = |\vec{P_0 A}|$.

Step-by-Step Solution

Step 1: Represent points P and Q parametrically. We are given two lines and a line segment PQ connecting them.

• The first line is $x = y+2 = z$. We can write this in symmetric form as $\frac{x}{1} = \frac{y+2}{1} = \frac{z}{1}$. Let a general point P on this line be $(t, t-2, t)$ for some parameter $t$.

• The second line is $x+2 = 2y = 2z$. We can write this in symmetric form as $\frac{x+2}{2} = \frac{y}{1} = \frac{z}{1}$. Let a general point Q on this line be $(s-2, s/2, s/2)$ for some parameter $s$. (We set $2y=2z=s$, so $y=s/2$, $z=s/2$, and $x+2=s \implies x=s-2$).

Step 2: Use the direction ratios of line PQ to find P and Q. The problem states that "A line with direction ratios $2,1,2$ meets the lines ... at P and Q". This implies that the line segment PQ has direction ratios proportional to $(2,1,2)$. The direction ratios of the vector $\vec{PQ}$ are $(x_Q-x_P, y_Q-y_P, z_Q-z_P)$. So, $\vec{PQ} = (s-2-t, s/2-(t-2), s/2-t) = (s-2-t, s/2-t+2, s/2-t)$. Since these direction ratios are proportional to $(2,1,2)$, we can write: $\frac{s-2-t}{2} = \frac{s/2-t+2}{1} = \frac{s/2-t}{2} = \lambda$ where $\lambda$ is a constant of proportionality.

From the first and third parts of the equation: $s-2-t = s/2-t$ $s-2 = s/2$ $s/2 = 2 \implies s = 4$

Now, substitute $s=4$ into the equations for $\lambda$: $\frac{4-2-t}{2} = \lambda \implies \frac{2-t}{2} = \lambda \quad (1)$ $\frac{4/2-t+2}{1} = \lambda \implies 2-t+2 = \lambda \implies 4-t = \lambda \quad (2)$

Equating (1) and (2): $\frac{2-t}{2} = 4-t$ $2-t = 2(4-t)$ $2-t = 8-2t$ $t = 6$

Now we can find the coordinates of P and Q:

• Point P: $(t, t-2, t) = (6, 6-2, 6) = (6,4,6)$.
• Point Q: $(s-2, s/2, s/2) = (4-2, 4/2, 4/2) = (2,2,2)$.

Let's verify the direction ratios of $\vec{PQ}$: $(2-6, 2-4, 2-6) = (-4, -2, -4)$. These are indeed proportional to $(2,1,2)$, with a proportionality constant of $-2$. So, the direction vector of line PQ can be taken as $\vec{d} = (2,1,2)$.

Step 3: Calculate the length of the perpendicular from point A(1,2,12) to line PQ. Let the given point be $A(1,2,12)$. Let $P_0$ be a point on the line PQ. We can use $P(6,4,6)$ as $P_0$. The direction vector of line PQ is $\vec{d} = (2,1,2)$.

First, form the vector $\vec{P_0 A}$: $\vec{P_0 A} = A - P_0 = (1-6, 2-4, 12-6) = (-5, -2, 6)$

Next, check if $\vec{P_0 A}$ is perpendicular to $\vec{d}$ by calculating their dot product: $\vec{P_0 A} \cdot \vec{d} = (-5)(2) + (-2)(1) + (6)(2)$ $= -10 - 2 + 12 = 0$ Since the dot product is 0, the vector $\vec{P_0 A}$ is perpendicular to the direction vector $\vec{d}$ of the line PQ. This means that $P_0$ (which is P) is the foot of the perpendicular from A to the line PQ.

Therefore, the length of the perpendicular $l$ is simply the distance between point A and point P. $l = |\vec{P_0 A}| = \sqrt{(-5)^2 + (-2)^2 + (6)^2}$ $l = \sqrt{25 + 4 + 36}$ $l = \sqrt{65}$

The question asks for $l^2$. $l^2 = (\sqrt{65})^2 = 65$

Common Mistakes & Tips

• Careful with parametric forms: Ensure correct representation of points on each line. A common mistake is to use the same parameter for both lines or incorrectly convert the line equations to parametric form.
• Direction Ratios vs. Direction Vector: Direction ratios are components of a vector parallel to the line. Any scalar multiple of $(a,b,c)$ can be used as the direction vector $\vec{d}$.
• Foot of Perpendicular: Always check if the vector from the chosen point on the line to the external point is perpendicular to the line's direction vector. If it is, the distance calculation simplifies significantly. Otherwise, use the cross product formula or find the foot of the perpendicular explicitly.

Summary

We first found the coordinates of points P and Q by setting up parametric equations for the two given lines and using the condition that the line segment PQ has direction ratios proportional to $(2,1,2)$. This yielded $P(6,4,6)$ and $Q(2,2,2)$. We then determined the direction vector of line PQ to be $\vec{d}=(2,1,2)$. Finally, we calculated the perpendicular distance from the point $A(1,2,12)$ to the line PQ. Since the vector $\vec{AP}$ was found to be perpendicular to the direction vector of line PQ, the distance was simply the length of $\vec{AP}$. The square of this distance, $l^2$, was calculated to be 65.

The final answer is $\boxed{65}$.