Can someone prove this why my answer is wrong?

Question

Find the linear equation of the plane through the point (1,2,3) and contains the line represented by the vector equation r(t)= <3t, 5-4t, 1-8t)

My answer was: 16x+14y+5z=59 but it is wrong. I need help showing me where I am wrong please

Answer 1

1. We are given that the point P (1,2,3) lies in the plane.
2. Since a plane is defined by 3 non-collinear points, let’s find a couple more points in the plane.
3. We know that the plane contains the line r(t) = (3t, 5-4t, 1-8t). So, we can find other points in the plane by substituting different values of t.
4. Substituting t=0 in r(t) gives us the point P1 (0,5,1).
5. Substituting t=1 in r(t) gives us the point P2 (3,1,-7).
6. Note that P does not lie on the line r(t) since we cannot find a value of t such that (1,2,3) = (3t, 5-4t, 1-8t).
7. Since P1 and P2 lie on r(t) and P does not, we conclude that P, P1, P2 are non-collinear and hence determine the plane.
8. Let’s calculate the vector PP1 which joins points P and P1. PP1 = 0i + 5j + k - (i + 2j + 3k) = -i + 3j - 2k
9. Similarly PP2 = (3i + j -7k) - (i + 2j + 3k) = 2i - j - 10k.
10. PP1 and PP2 lie in the plane, hence their cross-product N must be a normal (perpendicular) to the plane. N = PP1 x PP2 = -32i - 14j - 5k
11. Let X (x,y,z) be a point in the plane. Then, the vector PX joining points P and X lies in the plane and so is perpendicular to the vector N calculated above. So, the dot product PX.N = 0.
12. PX = (xi + yj + zk) - (i + 2j + 3k) = (x-1)i + (y-2)j + (z-3)k
13. Setting PX.N =0 gives us -32(x-1) - 14(y-2) - 5(z-3) = 0 which after some rearranging becomes 32x + 14y + 5z = 75.
14. So, the equation of the plane is 32x + 14y + 5z = 75.