Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has nonzero solutions and and at least one of the a,b,c is a proper fraction. System has solution such thatA. `x,y,z -= (1-2a^(2)):(1-2b^(2)):(1-2c^(2))`B. `x.y.z -= (1)/(1-2a^(2)):(1)/(1-2b^(2)):(1)/(1-2c^(2))`C. `x.y.z -= (a)/(1-a^(2)):(b)/(1-b^(2)):( c)/(1-c^(2))`D. `x.y.z -= sqrt(1-a^(2)):sqrt(1-b^(2)):sqrt(1-c^(2))`

Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has n

1.	Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has nonzero solutions and and at least one of the a,b,c is a proper fraction. System has solution such thatA. `x,y,z -= (1-2a^(2)):(1-2b^(2)):(1-2c^(2))`B. `x.y.z -= (1)/(1-2a^(2)):(1)/(1-2b^(2)):(1)/(1-2c^(2))`C. `x.y.z -= (a)/(1-a^(2)):(b)/(1-b^(2)):( c)/(1-c^(2))`D. `x.y.z -= sqrt(1-a^(2)):sqrt(1-b^(2)):sqrt(1-c^(2))`
Answer» Correct Answer - D The system of equation `-x+cy+bz=0` `cx-y+az=0` `bx+ay-z=0` has a nonzero solution if `Delta = \|{:(-1,,c,,b),(c,,-1,,a),(b,,a,,-1):}\|=0` then Clearly the system has infinitely many solutions .From (1) and (2) we have `(x)/(ac+b) =(y)/(bc+a)=(z)/(1-c^(2)) ` `" or " (x^(2))/((1-a^(2))(1-c^(2)))=(y^(2))/((1-b^(2))(1-c^(2)a)) =(z^(2))/((1-c^(2))^(2))`[From (4)] `" or " (x^(2))/(1-a^(2))=(y^(2))/(1-b^(2)) =(z^(2))/(1-c^(2))` from (5) we see that `1-a^(2),1-b^(2),1-c^(2)` are all positive or all negative .Given that one of a,b,c is proper fraction so `1-a^(2) gt ,1-b^(2) gt 0,1-c^(2) gt 0` which gives `a^(2) +b^(2)+c^(2) lt 3` using (4) and (6) we get `1lt 3+2 abc` `" or " abc gt -1`