Let $\overrightarrow a=\large\frac{1}{3}$$(\hat i-2\hat j+2\hat k),\:\overrightarrow b=\large\frac{1}{5}$$(-4i-3\hat k)$
$\Rightarrow\:|\overrightarrow a|=|\overrightarrow b|=1$
Let $\overrightarrow r=x\hat i+y\hat j+z\hat k$ make equal angle with $\overrightarrow a,\overrightarrow b\:and\:\hat j$
$\Rightarrow \overrightarrow r.\overrightarrow a=\overrightarrow r.\overrightarrow b=\overrightarrow r.\hat j$
$\Rightarrow\:\large\frac{x-2y+2z}{3}=\frac{-4x-3z}{5}=y$
$\Rightarrow\:x-5y+2z=0,\:4x+5y+3z=0\:and\:17x-10y+19z=0$
solving these three equations we get one of the values
$x=5,y=-1,z=-5$
$\therefore\:$The required vector $ = 5\hat i-\hat j-5\hat k$