To find the equation of the plane with a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\) and a point \(\displaystyle (x_0,y_0,z_0)\), we use the formula

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Using the information from the problem statement, we get 

\(\displaystyle 7(x-1)+5(y-3)+2(z-5)=0\)

This simplifies to \(\displaystyle 7x+5y+2z=32\)

To find the equation of the plane with a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\) and a point \(\displaystyle (x_0,y_0,z_0)\), we use the formula

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Using the information from the problem statement, we get 

\(\displaystyle 2(x-3)-5(y-7)+6(z-9)=0\)

This simplifies to \(\displaystyle 2x-5y+6z=25\)

To find the equation of a plane with a point \(\displaystyle (x_0,y_0,z_0)\) and a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the formula

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Using the information from the problem statement, we get

\(\displaystyle 4(x-3)+2(y+6)+1(z-7)=0\)

Simplifying, we then get

\(\displaystyle 4x+2y+z=7\)

To find the equation of a plane with a point \(\displaystyle (x_0,y_0,z_0)\) and a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the formula

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Using the information from the problem statement, we get

\(\displaystyle 4(x-5)+5(y+1)+5(z-2)=0\)

Simplifying, we then get

\(\displaystyle 4x+5y+5z=25\)

To find the equation of a plane, we need the normal vector and a point on the plane. The normal vector is found by taking the cross product of two vectors on the plane.

To find two vectors, simply find the difference between terminal and initial points:

\(\displaystyle \vec{AB}=\left \langle -1, 1, -9\right \rangle\), \(\displaystyle \vec{BC}=\left \langle -2, 1, -8 \right \rangle\)

Now, we can write the determinant in order to take the cross product of the two vectors:

\(\displaystyle \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ -1&1 &-9 \\ -2&1 &-8 \end{vmatrix}\)

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

\(\displaystyle -8\hat{i}-\hat{k}+18\hat{j}-(-2\hat{k}-9\hat{i}+8\hat{j})=\hat{i}+10\hat{j}+\hat{k}=\left \langle 1, 10, 1\right \rangle\)

Plugging this into the formula

\(\displaystyle a(x-x_0)+b(y-y_0)+c(z-z_0)=0\), where \(\displaystyle \vec{n}=\left \langle a, b, c\right \rangle\) and \(\displaystyle (x_0, y_0, z_0)\) is any of the given points (for example \(\displaystyle (1, 2 ,3)\)), we get

\(\displaystyle 1(x-1)+10(y-2)+1(z-3)=0\)

which simplified becomes

\(\displaystyle x+10y+z=24\)

To find the equation of a plane with a point \(\displaystyle (x_0,y_0,z_0)\) and normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the following equation:

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Plugging in the information from the problem statement, we get

\(\displaystyle 4(x-3)+4(y-0)+1(z-1)=0\)

Isolating the variables to one side gets us

\(\displaystyle 4x+4y+z=13\)

To find the equation of a plane with a point \(\displaystyle (x_0,y_0,z_0)\) and normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the following equation:

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Plugging in the information from the problem statement, we get

\(\displaystyle 2(x-2)+5(y-1)+2(z-5)=0\)

Isolating the variables to one side gets us

\(\displaystyle 2x+5y+2z=19\)

To find the equation of the plane containing a point \(\displaystyle (x_0,y_0,z_0)\) and a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the formula:

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Plugging in the known values and solving, we get

\(\displaystyle 1(x-3)+2(y+6)+1(z-7)=0\)

Simplifying, we get

\(\displaystyle x+2y+z=-2\)

To find the equation of the plane containing a point \(\displaystyle (x_0,y_0,z_0)\) and a normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the formula:

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Plugging in the known values and solving, we get

\(\displaystyle 3(x-5)+1(y-4)+5(z-1)=0\)

Simplifying, we get

\(\displaystyle 3x+y+5z=24\)

To find the equation of a plane given a point on the plane \(\displaystyle (x_0,y_0,z_0)\) and a normal vector to the plane \(\displaystyle n=\left \langle A,B,C\right \rangle\), we use the following equation

\(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)

Plugging in the information from the problem statement, we get

\(\displaystyle 8(x-3)+4(y+4)-5(z-3)=0\)

Rearranging, we get

\(\displaystyle 8x+4y-5z=-7\)