## Calculus (3rd Edition)

$$\langle 3,1,2 \rangle, \langle 6,0,4 \rangle.$$ (Other solutions are possible.)
Assume that the vector $\langle a, b,c \rangle$ is orthogonal to $\langle 2,0,-3 \rangle$, then we have $$\langle a,b,c \rangle \cdot \langle 2,0,-3\rangle=0\Longrightarrow 2a+0-3c=0\Longrightarrow a=\frac{3}{2}c.$$ We can pick any vector $\langle a,b,c \rangle$ that satisfies the above equation. Note that here, $b$ can be any value. Hence we can choose two vectors as follows $$\langle 3,1,2 \rangle, \langle 6,0,4 \rangle.$$ (Other solutions are possible.)