Consider the line $\frac{x-2}{2}=\frac{2y-5}{-3},z=-1$. Then which of the following is/are true?

(A) It has Direction ratios (2, -3, -1)
(B) It has Direction cosines $(\frac{4}{5},\frac{-3}{5},\frac{-1}{5})$
(C) It has Direction ratios $(2,\frac{-3}{2},0)$
(D) It has Direction cosines $(\frac{4}{5},\frac{-3}{5},0)$

Choose the correct answer from the options given below:

(C) and (D) only

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.