If L, M and N are natural number, then value of `LM=M+N=` ?
(A) `L+N=8+M`
(B) `M^(2)=(N^(2))/(L+1)`
(C) `M=L+2`

If the vectors vec a, vec b, vec c are non-coplanar and l,m,n are distinct real numbers, then [(l vec a + m vec b + n vec c) (l vec b + m vec c + n vec a) (l vec c + m vec a + n vec b)] = 0, implies (A) lm+mn+nl = 0 (B) l+m+n = 0 (C) l^2 + m^2 + n^2 = 0

If the vectors vec a, vec b, vec c are non-coplanar and l,m,n are distinct real numbers, then [( l vec a + m vec b + n vec c) (l vec b + m vec c + n vec a) (l vec c + m vec a + n vec b )] = 0, implies (A) lm+mn+nl = 0 (B) l+m+n = 0 (C) l^2 + m^2 + n^2 = 0

If the vectors vec a, vec b, vec c are non-coplanar and l,m,n are distinct real numbers, then [( l vec a + m vec b + n vec c) (l vec b + m vec c + n vec a) (l vec c + m vec a + n vec b )] = 0, implies (A) lm+mn+nl = 0 (B) l+m+n = 0 (C) l^2 + m^2 + n^2 = 0

If the vectors vec a, vec b, vec c are non-coplanar and l,m,n are distinct real numbers, then [( l vec a + m vec b + n vec c) (l vec b + m vec c + n vec a) (l vec c + m vec a + n vec b )] = 0, implies (A) lm+mn+nl = 0 (B) l+m+n = 0 (C) l^2 + m^2 + n^2 = 0

If L+M+N=0 and L,M,N are rationals the roots of the equation (M+N-L)x^(2)+(N+L-M)x+(L+M-N)=0 are

If l, m and n are real numbers such that l^(2)+m^(2)+n^(2)=0 , then |(1+l^(2),lm,l n),(lm,1+m^(2),mn),(l n,mn,1+n^(2))| is equal to a)0 b)1 c) l+m+n+2 d) 2(l+m+n)+3

Explain giving reasons which of the following sets of quantum number are not possible (a ) n=0, l =0 m_(l) = 0, m_(s ) =+ (1)/(2) ( b) n=1 , l = 0 m_(l) = 0, m_(s ) = - (1)/(2) ( c) n=1 , l = 1, m_(l ) = 0, m_(s ) = + (1)/(2) (d ) n= 2, l = 1, m_(l ) = 0, m_(s ) = (1) /(2) ( e) n=3, l = 3, m_(l) = 3, m_(s ) = + (1)/(2) (f ) n=3, l = 1, m_(l) = 0, m_(s) l = + (1)/(2)

IF l,m,n are rational the roots of (m +n) x^2 -(l+m+n)x+l =0 are

If l, m, n denote the side of a pedal triangle, then (l)/(a ^(2))+(m)/(b^(2))+(n)/(c ^(2)) is equal to