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 \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32