# Equation of plane

I have to find the equation of the plane that contains the line of intersection of x - y + 2y + 5 and 2x + 3y - z - 1 and is parallel to the line segment with normal vector [1, 2, -1].

I thought that I could go find the dot product of the normal vector I have [1, 2, -1] and the A B and C values for one of the two above equations.

x - y + 2y + 5 + k(2x + 3y - z - 1)=0
I rearrange it to get x(1 + 2k) + y(-1 + 3k) + z(2-k) + (5- k)=0. Now how can I find the value of k????

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I have to find the equation of the plane that contains the line of intersection of x - y + 2y + 5 and 2x + 3y - z - 1 and is parallel to the line segment with normal vector [1, 2, -1].

I thought that I could go find the dot product of the normal vector I have [1, 2, -1] and the A B and C values for one of the two above equations.

x - y + 2y + 5 + k(2x + 3y - z - 1)=0
I rearrange it to get x(1 + 2k) + y(-1 + 3k) + z(2-k) + (5- k)=0. Now how can I find the value of k????
I'm not sure yet about this argument, since I'm a bit unclear about what the problem intends by saying the plane is supposed to be "parallel to a line segment with normal vector <1, 2, -1>. It seems to me they could just say the plane you want has that as a normal vector. This would tell us that the coefficients for the plane you seek should be multiples of 1, 2, and -1.

As for your question, just pick some point that satisfies the equation for the plane

x - y + 2z + 5 = 0 .

Take any values you like for x and y, say, and solve that equation for z. Bingo! -- you have a point in that plane. Pick values for any two of the variables (even 0 and 0) and solve for the remaining variable -- it's a point in the plane. Putting those coordinates into your equation leaves you with one equation in one unknown. No matter what you use, you'll still get the same value for k.