Question

A line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ respectively at the points $P$ and $Q$. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$, then $l^2$ is

Answer 1

Step 1: Find Points $P$ and $Q$

Let $P(t, t - 2, t)$ and $Q(2s - 2, s, s)$ be the points where the line with direction ratios $2, 1, 2$ meets the given lines.

Step 2: Set Up Equations for Direction Ratios of $PQ$

The direction ratios (D.R.) of $PQ$ are $2, 1, 2$. Equating components:

$\frac{2s - 2 - t}{2} = \frac{s - t + 2}{1} = \frac{s - t}{2}$

Solving these equations, we find $t = 6$ and $s = 2$.

Step 3: Determine Coordinates of $P$ and $Q$

Substitute $t = 6$: $P(6, 4, 6)$. Substitute $s = 2$: $Q(2, 2, 2)$.

Step 4: Equation of Line $PQ$

The line $PQ$ can be written as:

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

Step 5: Find the Foot of Perpendicular $F$ from $A(1, 2, 12)$ to $PQ$

Let $F(2\lambda + 2, \lambda + 2, 2\lambda + 2)$ be the foot of the perpendicular. Since $\overrightarrow{AF} \cdot \overrightarrow{PQ} = 0$, solving gives $\lambda = 2$.

Step 6: Calculate $AF$

The coordinates of $F$ are $(6, 4, 6)$. Distance $AF$ is given by:

$AF = \sqrt{(6 - 1)^2 + (4 - 2)^2 + (6 - 12)^2} = \sqrt{65}$

Step 7: Find $l^2$

$l^2 = 65$

So, the correct answer is: 65