1. The problem statement, all variables and given/known data The two lines are in symmetric equations. L1 is (x-1)/2=(y+1)/3=-z. L2 is (x+1)/2=(y-1)/3=3-z. Find the equation of the plane containing the lines. The point on L1 is (1,-1,0) with slope of (2,3,-1) The point on L2 is (-1,1,3) with slope of (2,3,-1) 2. Relevant equations a(x-x0)+b(y-y0)+c(z-z0)=0 Normal vector =V(P1P2)XSlope 3. The attempt at a solution With same slope, you know the lines are parallel. So the vector from P1 to P2 is (-2,2,3) Cross product of vector P1 to P2 and the slope is (-11,-8,4) My thought was that the normal vector has the value of (a,b,c) in the equation for plane. I plugged in the values from the normal vector along with P1 to get: -11(x-1)-8(y+1)+4z=0. According to the solutions sheet, the answer is 11(x-1)-4(y+1)+10z=0. I got the x0,y0,z0 correct but why are the a, b, and c values wrong?