If vec{a}, vec{b}, text{ and } vec{c} are non | vedclass

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(B) Let the points be $A = 2 \vec{a}+3 \vec{b}-\vec{c}$,$B = 3 \vec{a}+4 \vec{b}-2 \vec{c}$,$C = \vec{a}-2 \vec{b}+3 \vec{c}$,and $D = \vec{a}-6 \vec{

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$\vec{a}+2 \vec{b}$

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(B) Let the points be $A = 2 \vec{a}+3 \vec{b}-\vec{c}$,$B = 3 \vec{a}+4 \vec{b}-2 \vec{c}$,$C = \vec{a}-2 \vec{b}+3 \vec{c}$,and $D = \vec{a}-6 \vec{b}+6 \vec{c}$.The equation of the line passing through $A$ and $B$ is given by $\vec{r} = A + t(B-A)$:$\vec{r} = (2 \vec{a}+3 \vec{b}-\vec{c}) + t((3 \vec{a}+4 \vec{b}-2 \vec{c}) - (2 \vec{a}+3 \vec{b}-\vec{c}))$$\vec{r} = (2+t) \vec{a} + (3+t) \vec{b} + (-1-t) \vec{c} \quad (i)$The equation of the line passing through $C$ and $D$ is given by $\vec{r} = C + s(D-C)$:$\vec{r} = (\vec{a}-2 \vec{b}+3 \vec{c}) + s((\vec{a}-6 \vec{b}+6 \vec{c}) - (\vec{a}-2 \vec{b}+3 \vec{c}))$$\vec{r} = \vec{a} + (-2-4s) \vec{b} + (3+3s) \vec{c} \quad (ii)$Comparing the coefficients of $\vec{a}, \vec{b}, 	ext{ and } \vec{c}$ in $(i)$ and $(ii)$:$2+t = 1 \implies t = -1$$3+t = -2-4s$$-1-t = 3+3s$Substituting $t = -1$ into the second equation:$3-1 = -2-4s \implies 2 = -2-4s \implies 4s = -4 \implies s = -1$Check consistency with the third equation:$-1 - (-1) = 3 + 3(-1) \implies 0 = 0$. The system is consistent.Substituting $t = -1$ into $(i)$:$\vec{r} = (2-1) \vec{a} + (3-1) \vec{b} + (-1-(-1)) \vec{c} = \vec{a} + 2 \vec{b}$.

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