Consider the line (x-2)/2=(2y-5)/(-3),z=-1. Then which of the followi…

The correct answer is Option (3) → (C) and (D) only

Given line: $\frac{x - 2}{2} = \frac{2y - 5}{-3},\ z = -1$

Rewrite the second equation:

$\frac{2y - 5}{-3} \Rightarrow \frac{y - \frac{5}{2}}{-\frac{3}{2}}$

Thus, the symmetric form is:

$\frac{x - 2}{2} = \frac{y - \frac{5}{2}}{-\frac{3}{2}},\ z = -1$

This is a line parallel to the vector:

Direction ratios = $(2,\ -\frac{3}{2},\ 0)$

Hence, (C) is correct.

Now compute direction cosines:

Let $\vec{v} = \langle 2,\ -\frac{3}{2},\ 0 \rangle$

$|\vec{v}| = \sqrt{2^2 + \left(-\frac{3}{2}\right)^2 + 0^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$

Direction cosines:

$\left( \frac{2}{\frac{5}{2}},\ \frac{-\frac{3}{2}}{\frac{5}{2}},\ 0 \right) = \left( \frac{4}{5},\ -\frac{3}{5},\ 0 \right)$

Hence, (D) is correct.