Let a, b and c be three non-coplanar vectors. | vedclass

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(B) The equation of the line passing through points $A(a+2b-5c)$ and $B(-a-2b-3c)$ is given by $r = A + \lambda_1(B-A)$.$r = (a+2b-5c) + \lambda_1(-2a

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$r=3a+6b-c+\mu(a+2b+c)$

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(B) The equation of the line passing through points $A(a+2b-5c)$ and $B(-a-2b-3c)$ is given by $r = A + \lambda_1(B-A)$.$r = (a+2b-5c) + \lambda_1(-2a-4b+2c) = (a+2b-5c) + \lambda_1'(a+2b-c)$ where $\lambda_1' = -2\lambda_1$.For simplicity,let $r = (a+2b-5c) + \lambda_1(a+2b-c)$ $(i)$.The equation of the line passing through points $C(-4c)$ and $D(6a-4b+4c)$ is $r = C + \lambda_2(D-C)$.$r = -4c + \lambda_2(6a-4b+8c) = -4c + 2\lambda_2(3a-2b+4c)$ (ii).Equating $(i)$ and (ii) for the intersection point:$(1+\lambda_1)a + (2+2\lambda_1)b + (-5-\lambda_1)c = (6\lambda_2)a + (-4\lambda_2)b + (-4+8\lambda_2)c$.Comparing coefficients of $a, b, c$:$1+\lambda_1 = 6\lambda_2$,$2+2\lambda_1 = -4\lambda_2$,$-5-\lambda_1 = -4+8\lambda_2$.From the first two equations: $2(1+\lambda_1) = 2+2\lambda_1 = -4\lambda_2$,so $2(6\lambda_2) = -4\lambda_2 \implies 16\lambda_2 = 0 \implies \lambda_2 = 0$.Then $\lambda_1 = -1$. The intersection point is $r = -4c + 0 = -4c$.Checking the options,for $\mu = -3$,option $B$ gives $r = (3a+6b-c) - 3(a+2b+c) = 3a+6b-c-3a-6b-3c = -4c$. Thus,the line in option $B$ passes through the intersection point.

Let $a, b$ and $c$ be three non-coplanar vectors. The vector equation of a line which passes through the point of intersection of two lines,one joining the points $a+2b-5c$ and $-a-2b-3c$,and the other joining the points $-4c$ and $6a-4b+4c$,is

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