Find the equation of plane[ If 2x+3y+4z=5 and 4x-5y+6z=7 passing through (1,2,-3)]
Since,
W.K.T
eq(1)+A[eq(2)] = 0
2x+3y+4z-5+A(4x-5y+6z-7) = 0
Passing through (1,2,-3) therefore x = 1 , y = 2, and z = -3:
2+6-12-5+A(4-10-18-7) = 0
-9-31A=0
A = -9/31
So equation is :
2x+3y+4z-5 -9/31(4x-5y+6z-7)=0
(2x-36x/31)+(3y+45y/31)+(4z-54z/31) -(5-63/31)
62x-36x+93y+45y+124z-54z-(155-63) = 0
26x+138y+70z-92 = 0
So the equation of plane is 26x+138y+70z-92=0.
Thank you!❤️