Vector equation of a plane passing through $r=(1,1,-2)$, $s=(3,0,1)$, $p=(1,1,1)$

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For this question I did the cross product so I found $rs$ then $rp$ and using those I did the cross product method to find $(a,b,c)$. After getting $a$, $b$, $c$ which was $(-3,-6,0)$. I plugged it into $a(x-x_0)+b(y-y_0)+c(z-z_0)$. This gave me a final equation of $3x+6y=9$.

The answer however is completely different where they said the equation is $(1,1,-2)+ \lambda(2,-1,3)+\mu(0,0,3)$ .

How would I know if my answer was correct from this equation and also how will I know when to do the cross product and when to not do it.

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If you set $(x,y,z)=(1,1,-2)+ \lambda(2,-1,3)+\mu(0,0,3)=(1+2\lambda,1-\lambda,-2+3\lambda+3\mu)$, then

$3x+6y=3(1+2\lambda)+6(1-\lambda)=9$

The two forms essentially represent the same plane.

(Since only $z$ depends on $\mu$, $z$ can be any real number.)