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VECTOR PROBLEM The cross product of the normal vectors to two planes is a vector that points in the direction of the line of intersection of the planes. Find a particular equation of the plane containing (-3, 6, 5) and normal to the line of intersection of the planes 3x + 5y + 4z = -13 and 6x - 2y + 7z = 8 Can someone guide me through it?

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Since no one is responding, let me start with a simpler question. Would the cross product be a vector that is perpendicular to the line it is pointing towards? If so, it would be a bit simpler...
Well finding the cross product is pretty simple. (3,5,4)x(6,-2,7)=(43,3,-36) This is the direction the line of intersection is pointing in. I'm just confused about what they mean by a plane that is normal to the line of intersection. Does the plane pass through the line? Is the normal perpendicular to the line?
Here's a cute little graph I made, it might help :P Blue lines are normals to planes and red line is cross product.

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Yes, a normal when talking about vectors is defined as a vector (or something else) that is perpendicular to what it is normal to.
OK I get it, the cross product will the normal of your new plane. So this is easy. Your new plane looks like: 43x+3y-36z=d Plug in your point (x,y,z) and solve for d!
OH! I get it now...I read the question incorrectly... -_- I thought it said that it pointed in the direction of the line, not the direction the line would be now since the normal and plane have the same coefficients, and because the line of intersection and the vector are parallel, I can just plug in the point on the plane to finish the problem?
Yeah, exactly. You're basically looking for a plane that is normal to the other two planes, they're just making it overly complicated by talking about the line of intersection.
Thank you!

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