If $\vec r=3\hat i+2\hat j-5\hat k, \vec a=2\hat i-\hat j+\hat k, \vec b=\hat i+3\hat j-2\hat k$ and $\vec c=-2\hat i+\hat j-3\hat k$ such that $\vec r=x\vec a+y\vec b+z\vec c$ then

$y,\frac{x}{2},z$ are in AP

We have,

$\vec r=x\vec a+y\vec b+z\vec c$

$⇒3\hat i+2\hat j-5\hat k=x(2\hat i-\hat j+\hat k)+y(\hat i+3\hat j-2\hat k)+z(-2\hat i+\hat j-3\hat k)$

$⇒3=2x+y-2z, 2=-x+3y+z, -5=x-2y-3z$

$⇒x=3, y=1, z=2$

Clearly, $x = y + z⇒y,\frac{x}{2},z$ are in AP.