Question

Let the line $L$ intersect the lines
$x - 2 = -y = z - 1$, $\quad 2(x + 1) = 2(y - 1) = z + 1$
and be parallel to the line
$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$.
Then which of the following points lies on $L$?

• A. \(\left( \frac{1}{3}, 1, 1 \right)\)
• B. \(\left( \frac{1}{3}, 1, -1 \right)\)
• C. \(\left( \frac{1}{3}, -1, -1 \right)\)
• D. \(\left( \frac{1}{3}, -1, 1 \right)\)

Answer 1

The Correct Option is B

Given the two lines: 

\( L_1 : \frac{x - 2}{1} = \frac{y}{-1} = \frac{z - 1}{1} = \lambda \)

\( L_2 : \frac{x + 1}{2} = \frac{y - 1}{-1} = \frac{z + 1}{1} = \mu \)

The direction ratios (d.r.) of the line \( MN \) will be:

\( \langle 3 + \lambda - \frac{\mu}{2}, -1 - \lambda - \frac{\mu}{2}, 2 + \lambda - \mu \rangle \)

and it will be proportional to \( \langle 3, 1, 2 \rangle \).

Hence,

\( \frac{3 + \lambda - \frac{\mu}{2}}{3} = \frac{-1 - \lambda - \frac{\mu}{2}}{1} = \frac{2 + \lambda - \mu}{2} \)

From the first two ratios:

\( 4\lambda + \mu = -6 \)

From the second and third ratios:

\( 4 + 3\lambda = 0 \)

Solving these equations:

\( \lambda = -\frac{4}{3}, \quad \mu = -\frac{2}{3} \)

Therefore, the coordinates of point \( M \) are:

\( M \left( \frac{4}{3}, \frac{4}{3}, -\frac{1}{3} \right) \)

and the equation of the required line is:

\( \frac{x}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k \)

So, any point on this line will be:

\( \left( \frac{2}{3} + 3k, \frac{4}{3} + k, -\frac{1}{3} + 2k \right) \)

Substituting \( k = -\frac{1}{3} \):

\( \text{Point on the line} = \left( -\frac{1}{3}, 1, -1 \right) \)

Answer 2

The given line \( L \) is parallel to the line:

\[ \frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}, \]

which implies the direction ratios of \( L \) are \( 3 : 1 : 2 \).

Assume that \( L \) intersects the first line \( x - 2 = -y = z - 1 \) at some point \( P \). From the equation of the line:

\[ x - 2 = -y = z - 1 = k. \]

This gives:

\[ x = 2 + 3k, \quad y = -k, \quad z = 1 + k. \]

Next, assume \( L \) also intersects the second line \( 2(x + 1) = 2(y - 1) = z + 1 \) at some point \( Q \). From the equation of the line:

\[ 2(x + 1) = 2(y - 1) = z + 1 = m. \]

This gives:

\[ x = m - 1, \quad y = \frac{m}{2} + 1, \quad z = m - 1. \]

Now, the line \( L \) passes through both points \( P \) and \( Q \), and we know it is parallel to the direction ratios \( 3 : 1 : 2 \). Substituting the parametric forms into the equations of the line \( L \) and solving for \( k \) and \( m \), we determine the valid points that lie on \( L \).

After solving, it is found that the point:

\[ \left( -\frac{1}{3}, 1, -1 \right) \]

satisfies the equation of \( L \).

Thus, the correct answer is option (2).