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 $L$?
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).
Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:
Let the shortest distance between the lines $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha + \beta$ is
Match List-I with List-II: List-I