It is given that $\vec{t_1}=2\hat i-\hat j+\hat k$ and $\vec{t_2}=\hat i+3\hat j-2\hat k$, $\vec{t_3}=-2\hat i+\hat j-3\hat k$ and $\vec{t_4}=3\hat i+2\hat j-5\hat k$. If $\vec{t_4}=p$  $\vec{t_1}+q\vec{t_2}+r\vec{t_3}$, then

p = q + r

We have $3\hat i+2\hat j-5\hat k= p(2\hat i-\hat j+\hat k) + q(\hat i+3\hat j-2\hat k) + r(-2\hat i+\hat j-3\hat k)$

⇒ 2p + q – 2r = 3, …….. (1)

– p + 3q +r =2,

p – 2q – 3r = –5.

Adding all the equations, we get

2p +2q = 4r i.e. p + q = 2r.

Equation (1) gives p = 3 ⇒ q = 1, r = 2.

With these values of q, p, r, only (B) is the correct answer.