The line joining the points 6a - 4b + 4c and | vedclass

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(D) Let the points be $P_1 = 6a - 4b + 4c$,$P_2 = -4c$,$P_3 = -a - 2b - 3c$,and $P_4 = a + 2b - 5c$. The equation of the line passing through $P_1$ an

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None of these

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(D) Let the points be $P_1 = 6a - 4b + 4c$,$P_2 = -4c$,$P_3 = -a - 2b - 3c$,and $P_4 = a + 2b - 5c$. The equation of the line passing through $P_1$ and $P_2$ is $r = P_1 + m(P_2 - P_1) = 6a - 4b + 4c + m(-6a - 4b - 8c)$ ..... $(i)$. The equation of the line passing through $P_3$ and $P_4$ is $r = P_3 + n(P_4 - P_3) = -a - 2b - 3c + n(2a + 4b - 2c)$ ..... $(ii)$. For the point of intersection,the equations $(i)$ and $(ii)$ must yield the same position vector $r$. Equating the coefficients of $a, b,$ and $c$ from $(i)$ and $(ii)$: $6 - 6m = -1 + 2n \implies 6m + 2n = 7$ $-4 - 4m = -2 + 4n \implies 4m + 4n = -2 \implies 2m + 2n = -1$ $4 - 8m = -3 - 2c \implies 8m - 2n = 7$ Subtracting the second equation from the first: $(6m + 2n) - (2m + 2n) = 7 - (-1) \implies 4m = 8 \implies m = 2$. Substituting $m = 2$ into $2m + 2n = -1$: $4 + 2n = -1 \implies 2n = -5 \implies n = -2.5$. Checking these values in the third equation: $8(2) - 2(-2.5) = 16 + 5 = 21 
eq 7$. Since the system of equations is inconsistent,the lines do not intersect. Therefore,the correct option is $(d)$.

The line joining the points $6a - 4b + 4c$ and $-4c$ and the line joining the points $-a - 2b - 3c$ and $a + 2b - 5c$ intersect at:

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