Equation of a plan through three points? Can someone explain how they got the final answer when they cross product? (10i + 8j + 10k) 

 A: $10{\hat i}+8{\hat j}+10{\hat k}$ is not the final answer but rather represents the normal vector to the required plane. The "table" you are referring to is a matrix whose determinant give the cross product. While this is the most common way to find the determinant, you could also just simply cross them like this-
$$\bigr(-{\hat i}+5{\hat j}-3{\hat k}\bigr)\times \bigr(-2{\hat i}+0{\hat j}+2{\hat k}\bigr)$$
$$2\bigr({\hat i}\times {\hat i}\bigr)+0\bigr({\hat i}\times {\hat j}\bigr)-2\bigr({\hat i}\times {\hat k}\bigr)-10\bigr({\hat j}\times {\hat i}\bigr)+0\bigr({\hat j}\times {\hat j}\bigr)+10\bigr({\hat j}\times {\hat k}\bigr)+6\bigr({\hat k}\times {\hat i}\bigr)+0\bigr({\hat k}\times {\hat j}\bigr)-6\bigr({\hat k}\times {\hat k}\bigr)$$
$$2{\hat j}+10{\hat k}+10{\hat i}+6{\hat j}$$
$$10{\hat i}+8{\hat j}+10{\hat k}$$
Evidently, the determinant method is easier to do as well as much faster.
To directly get the equation of the plane, it would be best to solve this determinant instead-
$$
    \begin{vmatrix}
    x-2 & y+1 & z-3 \\
    -1 & 5 & -3 \\
    -2 & 0 & 2 
    \end{vmatrix}=0
$$
