The lines $¯ ¯ ¯ r = (6 - 6 s) ¯ ¯ ¯ a + (4 s - 4) ¯ ¯ b + (4 - 8 s) ¯ ¯ c$ and $¯ ¯ ¯ r = (2 t - 1) ¯ ¯ ¯ a + (4 t - 2) ¯ ¯ b - 2 (t + 3) ¯ ¯ c$ intersect at

4 ¯ ¯ c

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- 4 ¯ ¯ c

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3 ¯ ¯ c

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- 2 ¯ ¯ c

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Solution

The correct option is B $- 4 ¯ ¯ c$
Since vectors have intersection point so equate
$(6 - 6 s) \to a + (4 s - 4) \to b + (4 - 8 s) \to c = (2 t - 1) \to a + (4 t - 2) \to b - 2 (t + 3) \to c$
$6 - 6 s = 2 t - 1$ (1)
$4 s - 4 = 4 t - 2$ (2)
$2 \times (1) - (2)$
$s = 1, t = \frac{1}{2}$
so $\to r = - 4 \to c$

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