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

Correct Answer: 65

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