cross product

Suppose there are solutions for $(1,2,3) \times \vec{x} = (-1,8,a)$ , then $(1,2,3)\cdot (-1,8,a)=0$ . This means that $-1+16+3a=0$ , and hence $a=-5$ .

We check that $(1,2,3) \times (0,-5,-8) = (-1,8,-5)$ .

Hence only $\boxed{\text{one}}$ such $a$ .

Let $\vec x=\hat i p+\hat j q+\hat k r$ . Then the given equation reduces to

$\begin {pmatrix} 0 & -3 & 2\\3 & 0 & -1\\-2 & 1 & 0\end {pmatrix}\begin {pmatrix} p\\q\\r\end {pmatrix}=\begin {pmatrix} -1 \\8\\a\end {pmatrix}$

The determinant of the coefficient matrix is

$\left |\begin {matrix} 0 & -3 & 2\\3 & 0 & -1\\-2 & 1 & 0\end {matrix}\right |=0$

Hence the value of the determinant

$\left |\begin {matrix} -1 & -3 & 2\\8 & 0 & -1\\a & 1 & 0\end {matrix}\right |$

must be zero. So $a=-5$ , and this is the only value of $a$ .

Therefore the answer is $\boxed 1$ .