In `DeltaABC` as shown, `XX_1=d_1 ; XX_2=d_2 ;XX_3 = d_3 and X` is the centre of the circumscribed circle around the `DeltaABC, a, b and c` as usual are the sides BC, CA and AB respectively .If `lambda(a/d_1+b/d_2+c/d_3)=(abc)/(d_1 d_2 d_3)` , then the value of `'lambda'` is equal to

Let f,g,h be the lengths of the perpendiculars from the circumcenter of the Delta ABC on the sides BC,CA and AB respectively.If (a)/(f)+(b)/(g)+(c)/(h)=lambda(abc)/(fgh), then the value of 'lambda' is (B) (1)/(2) (B) (1)/(2) (D) (1)/(2) (D) 2

If ( a times b)*(c times d)=(a*c)(b*d)+k(a.d)(b.c) then the value of k is

If a + 2b - c = 0 and a xx b + b xx c + c xx a = lambda (a xx b) , then the value of lambda is equal to a)5 b)4 c)2 d)-2

If D,E and F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC and lambda is scalar, such that vec(AD) + 2/3vec(BE)+1/3vec(CF)=lambdavec(AC) , then lambda is equal to

If D,E and F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC and lambda is scalar, such that vec(AD) + 2/3vec(BE)+1/3vec(CF)=lambdavec(AC) , then lambda is equal to

If D,E and F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC and lambda is scalar, such that vec(AD) + 2/3vec(BE)+1/3vec(CF)=lambdavec(AC) , then lambda is equal to

The points D, E, F are the midpoints of the sides bar(BC), bar(CA), bar(AB) of DeltaABC respectively. If A = (-2, 3), D = (1, -4), E = (-5, 2), then F =

In a Delta ABC, the distance of sides BC,AC and AB from circumcentre are d_(1),d_(2) and d_(3) respectively (where a,b&c are length of sides in usual notation If ((a)/(d_(1))+(b)/(d_(2))+(c)/(d_(3)))=k(abc)/(d_(1)d_(2)d_(3)), then the value of k' is (lambda)/(mu),lambda,mu in N and are comprime then

If the distance of the plane x - y + z + lambda = 0 from the point (1, 1, 1) is d_1 and the distance of this point from the origin is d_2 and d_2d_2 = 5 then find the value of lambda .