Answer:
Option A
Explanation:
The given line is $\frac{x-2}{2}=\frac{2y-5}{-3} = z+1,$
$\frac{x-2}{2} = \frac{y-\frac{5}{2}}{-\frac{3}{2}} = \frac{z+1}{0}$
This shows that the given line passes through the point (2,5/2,-1) and has direction ratios (2,-3/2,0).
Thus. given line passes through the point having Position vector
$\overrightarrow{a}$ = $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})$
$\overrightarrow{b}$ = $\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$
$\overrightarrow{r}$= $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$
Hence, p = 0