A primal approach

Solution : $y(z^2 - x^2) + x(y^2 - z^2) + z(x^2-y^2)=p$

Hence, $yz^2+zx^2+xy^2-xz^2-zy^2-yx^2=p$

Adding and subtracting $xyz$ we get:

$xyz-xz^2-zy^2+yz^2-yx^2+zx^2+xy^2-xyz=p$

So, $(z-x)(xy-xz-y^2+yz)=p$

and thus, $(z-x)(y-z)(x-y)=p$

Now pay attention that $y-x$ can be achieved through adding $y-z$ and $z-x$ .And since $p$ is a prime number and $x,y,z \in \mathbb {z}$ we can deduce the following:

1. $y-z$ can have 4 different values at most: $(p),(1),(-1),(-p)$

2. Same as the first deduction, $z-x$ can also have at most 4 values same as above.

Hence, there are 4 cases for $y-z$ and similarly 4 cases for $z-x$ .We investigate the cases for $y-z$ and divide them into 2 cases each containing 2 sub-cases.

1. $y-z=1 \lor -1$

$\Rightarrow z-x=-1 \lor 1 \lor p \lor -p$

$\Rightarrow y-z= {0 \lor 2 \lor p+1 \lor -p+1 \lor -2 \lor p-1 \lor-p-1} \space \space \space \space \text {an extra 0 is omitted}$

1. $y-z=p \lor -p$

$\Rightarrow z-x=-1 \lor 1 \space \space \space \space \text {(why?)}$

$\Rightarrow y-z=p-1 \lor p+1 \lor -p-1 \lor -p+1$

Note that the solutions found within the investigation of case no.2 are in advance counted in case no.1 solutions and thus they won't add any thing to the solutions from case no.1.

And the answer -as can be seen in case no.1- is 7.

Have fun!

Please correct me wherever I went wrong and leave feel free to express your opinions.