To find the equation of a plane given a point on the plane  and a normal vector to the plane , we use the following equation

Plugging in the information from the problem statement, we get

Rearranging, we get

To find the equation of a plane that contains a point  and a normal vector , we use the equation

Since we know the point on the plane as well as the normal vector (two parallel planes contain the same normal vector, so in this case it is , we can plug what we know into the equation

Rearranging, we get

To find the equation of a plane that contains a point  and a normal vector , we use the equation

Since we know the point on the plane as well as the normal vector (two parallel planes contain the same normal vector, so in this case it is , we can plug what we know into the equation

Rearranging, we get

The equation of a plane is given by

where  and  is any point on the plane. 

First, we must create two vectors out of the given points (by subtracting terminal and initial points):

, 

Now, we can write the determinant in order to take the cross product of the two vectors, which will give us the normal vector:

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:

Now that we have the normal vector, we can pick any point on the plane, and plug all of this into the formula above:

which simplified becomes

The equation of a plane is given by

where the normal vector is given by  and a point on the plane denoted 

To find the normal vector to the plane, we must take the cross product of the two vectors.

We must write the determinant in order to take the cross product:

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:

Now that we have the normal vector and a point on the plane, we plug everything into the equation:

which simplifies to

The equation of the plane parallel to two lines is found by taking the point, and the normal vector to both lines (which in turn is normal - perpendicular - to the plane in question), and plugging them into the equation

where , and the point on the plane .

We must write the determinant in order to take the cross product of the two vectors (which gives us the normal vector):

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:

We now have everything to plug into the equation.

We get

which simplified becomes

First, we need two vectors on the plane so we can form the normal vector to the plane. To do this, we take  and , to form the vectors  and .

Next we take the cross product of the vectors  and 

Using the formula for the cross product, we find that the normal vector is 

We then use the formula for a plane with a normal vector  and a point  

Using the point , we get

Simplifying, we get 

First, we need to find the normal vector to the plane, which is one by taking the cross product of the vectors that are parallel to the plane.

Using the formula for the cross product, we get that the normal vector is

Using the formula of a plane with a normal vector  and containing a point , we get

Plugging in what we have, we get

Simplifying, we get

If a plane is parallel to two vectors, then the vector perpendicular to those vectors must be perpendicular to the plane as well.

Using this fact, we take the cross product of the two vectors to get the normal vector, which is also the normal vector of the plane, .

The equation of a plane is given by

, where  is a point on the plane.

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

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:

Plugging this, and the point given, into our equation, we get

which simplifies to

The equation of a plane is given by

where the normal vector to the plane is  and a point on the plane . 

The plane is parallel to two vectors, so anything perpendicular to those vectors will be perpendicular to the plane. 

So, we must take the cross product of the two vectors to get the vector normal to them (and therefore normal to the plane).

We can write the determinant in order to take the cross product of the two vectors:

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:

Plugging in all of our known information into the equation above, we get

which simplifies to