Q.
If the equation of the plane passing through the point (−1,2,0) and parallel to the lines 3x=0y+1=−1z−2 and 1x−1=2y+1=−1z+1 is ax+by+cz=1, then the value of (a+b+c), is
Normal vector of the plane h=∣∣i^31j^02k^−1−1∣∣ h=2i^+2j^+6k^=2(i^+j^+3k^) ∴ Equation of plane 1(x+1)+1(y−2)+3(z−0)=0
P: x+y+3z=1
hence, (a+b+c)=1+1+3=5.