Equation of a line given one point and two planes

Saaady

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I've done a question similar to this, however this one has no complete equations i can solve for.
Determine the equation of the plane that passes through [FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT](1,3,8) and is perpendicular to the line of intersection of the planes [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]3x−2z+1=0 and [FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]4x+3y+7=0.
I know to take the cross product of the two normals to get my new direction vector, but im stuck at that point.
 
I've done a question similar to this, however this one has no complete equations i can solve for.
Determine the equation of the plane that passes through [FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT](1,3,8) and is perpendicular to the line of intersection of the planes [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]3x−2z+1=0 and [FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]4x+3y+7=0.
I know to take the cross product of the two normals to get my new direction vector, but im stuck at that point.

What is the equation of line of intersection of planes [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]3x−2z+1=0 and [FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]4x+3y+7=0 ?
 
What is the equation of line of intersection of planes [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]3x−2z+1=0 and [FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]4x+3y+7=0 ?

Yes so the question reads "determine the equation of the plane that passes through (1,3,8) and is perpendicular to the line of intersection of the pleanes 3x-2z+1=0 and 4x+3y+7=0"

I'm unsure how to find my new normal given i don't have complete equations for the planes, and also how to write the new equation.
 
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