The Total solution for NCERT class 6-12
x = 4, y тАУ 3, z = 2тЗТ┬а4x2┬а+ y2┬а+25z2┬а+ 4xy тАУ 10yz тАУ 20zx= (2x)2┬а+ (y)2┬а+ (5z)2┬а+ 2 x2 xx y-2 x y x 5z тАУ 2 x 5z x 2x= (2x + y- 5z)2= (2 x 4 + 3- 5 x 2)2= (8 + 3- 10)2= (11 тАУ 10)2= (1)2┬а= 1