The lines lx+my+n=0,mx+ny+l=0 and nx+ly+m=0 are concurrent if
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a
l+m−n=0
b
l−m−n=0
c
l−m+n=0
d
l2+m2+n2=lm+mn+nl
answer is D.
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Detailed Solution
Lines are concurrent if lmnmnlnlm=0⇒l+m+nmnl+m+nnll+m+nlm=0 C1→C1+C2+C3⇒(l+m+n)1 mn1nl1lm=0⇒(l+m+n)mn+nl+lm−l2−m2−n2=0⇒l+m+n=0;l2+m2+n2−lm−mn−nl=0∴l2+m2+n2=lm+mn+nl