यदि रैखिक समीकरणो के निकाय 2x+2y+3z=a 3x-y+5z=b x-3y+2z=c(जहाँ a,b,c शून्येतर वास्तविक संख्याएँ ...

2x+3y+3z=5 x-2y+z=-4 3x-y-2z=3

x+y+z=1 x-2y+3z=2 5x-3y+z=3

2x + 3y + 3z = 5, x-2y + z = -4,3x-y-2z = 3

2x + 3y-5z = 7, x + y + z = 6,3x-4y + 2z = 1, then x =

3x-4y+2z=-1 2x+3y+5z=7 x+z=2

If the system of linear equations 2x + 2y +3z =a 3x - y +5z =b x- 3y +2z =c where a, b, c are non-zero real numbers, has more than one solution, then

2x-3y+5z=16, 3x+ 2y-4z= -4, x + y - 2z =- 3.

Show that |a b c a+2x b+2y c+2z x y z|=0