write a scalar equation, given it's vector equation.

[x,y,z]= [3,7,-5] + s[1,2,-1] + t[5,-2,3]

( I remember my teacher saying that you have to cross s and t, and that will be my normal and thereafter I substitue the point to find the 'D' value of the scalar equation, but when I try it, it does not give me the answer? Is my method wrong?)

did you get a reduced cross-product of (1,-2,-3) ?

Here is an on-line cross-product calculator:
http://www.analyzemath.com/vector_calculators/vector_cross_product.html

so your equation would be x - 2y - 3z = D
sub in your given point and you got it.

Your method is on the right track, but there seems to be a misunderstanding in regards to the normal vector and the value of 'D'. Let me explain the correct method step by step.

The given vector equation is [x, y, z] = [3, 7, -5] + s[1, 2, -1] + t[5, -2, 3].

To convert this vector equation into a scalar equation, we need to find the normal vector, usually denoted as N, and a constant term, usually denoted as D.

1. Begin by finding the direction vectors, let's call them A and B.
A = [1, 2, -1]
B = [5, -2, 3]

2. Take the cross product of A and B to find the normal vector, N.
N = A x B = [(2 * 3) - (-1 * (-2)), (-1 * 5) - (1 * 3), (1 * (-2)) - (2 * 5)]
= [8, -8, -12]

3. Now, substitute the given point [3, 7, -5] into the scalar equation to find the value of D.
[x, y, z] = [3, 7, -5]
N · [x, y, z] = N · [3, 7, -5]
8x - 8y - 12z = 8(3) - 8(7) - 12(-5)
= 24 - 56 + 60
= 28

4. Finally, write the scalar equation using N and D:
8x - 8y - 12z = 28

So, the correct scalar equation for the given vector equation is 8x - 8y - 12z = 28.