Step 1: Direction ratios of the given line
The direction ratios of the line:
\[ \frac{3x - 11}{2} = \frac{3y - 11}{1} = \frac{3z - 19}{2} \]
are:
\[ \vec{d} = \langle 2, 1, 2 \rangle \]
Step 2: Representing point \( B \)
Let point \( B \) on the given line be:
\[ B = (1 + \lambda, 2 + \lambda, 3 + 2\lambda), \]
where \(\lambda\) is the parameter.
Step 3: Direction ratios of line \( AB \)
Point \( A = \left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right) \). The direction ratios of \( AB \) are:
\[ \text{D.R. of } AB = \left\langle \frac{3\lambda - 8}{3}, \frac{3\lambda - 5}{3}, \frac{6\lambda - 10}{3} \right\rangle \]
Step 4: Parallel condition
Since \( AB \) lies along the direction vector \(\vec{d} = \langle 2, 1, 2 \rangle\), we have:
\[ \frac{\frac{3\lambda - 8}{3}}{2} = \frac{\frac{3\lambda - 5}{3}}{1} = \frac{\frac{6\lambda - 10}{3}}{2} \]
Simplify the first ratio:
\[ \frac{3\lambda - 8}{3 \cdot 2} = \frac{3\lambda - 5}{3} \]
Cross-multiply:
\[ 3\lambda - 8 = 6\lambda - 10 \]
Solve for \(\lambda\):
\[ 3\lambda = 2 \implies \lambda = \frac{2}{3} \]
Step 5: Find \( B \)
Substitute \(\lambda = \frac{2}{3}\) into \( B = (1 + \lambda, 2 + \lambda, 3 + 2\lambda) \):
\[ B = \left(1 + \frac{2}{3}, 2 + \frac{2}{3}, 3 + 2 \cdot \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \]
Step 6: Find distance \( AB \)
The distance \( AB \) is given by:
\[ AB = \sqrt{\left(\frac{11}{3} - \frac{5}{3}\right)^2 + \left(\frac{11}{3} - \frac{8}{3}\right)^2 + \left(\frac{19}{3} - \frac{13}{3}\right)^2} \]
Simplify each term:
\[ AB = \sqrt{\left(\frac{6}{3}\right)^2 + \left(\frac{3}{3}\right)^2 + \left(\frac{6}{3}\right)^2} \]
\[ AB = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} \]
Thus:
\[ AB = 3 \]
\(Option (1) : \; 3\)