If A(1, 0, 2),B(2, 1, 0),C(2, -5, 3),and D(0, | vedclass

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(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is given by $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda

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$3$

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(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is given by $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Rightarrow \frac{x-1}{1} = \frac{y}{1} = \frac{z-2}{-2} = \lambda$. Thus,any point on $AB$ is $(\lambda+1, \lambda, -2\lambda+2)$. The equation of line $CD$ passing through $C(2, -5, 3)$ and $D(0, 3, 2)$ is given by $\frac{x-2}{0-2} = \frac{y+5}{3-(-5)} = \frac{z-3}{2-3} = \mu$ $\Rightarrow \frac{x-2}{-2} = \frac{y+5}{8} = \frac{z-3}{-1} = \mu$. Thus,any point on $CD$ is $(-2\mu+2, 8\mu-5, -\mu+3)$. For the intersection point $P$,we equate the coordinates: $1) \lambda+1 = -2\mu+2 \Rightarrow \lambda + 2\mu = 1$ $2) \lambda = 8\mu-5 \Rightarrow \lambda - 8\mu = -5$ Subtracting $(2)$ from $(1)$: $10\mu = 6 \Rightarrow \mu = \frac{3}{5}$. Substituting $\mu$ in $(1)$: $\lambda + 2(\frac{3}{5}) = 1 \Rightarrow \lambda = 1 - \frac{6}{5} = -\frac{1}{5}$. Wait,checking the intersection: $x = -\frac{1}{5} + 1 = \frac{4}{5}$,$y = -\frac{1}{5}$,$z = -2(-\frac{1}{5}) + 2 = \frac{12}{5}$. Using $\mu = \frac{3}{5}$: $x = -2(\frac{3}{5}) + 2 = \frac{4}{5}$,$y = 8(\frac{3}{5}) - 5 = -\frac{1}{5}$,$z = -(\frac{3}{5}) + 3 = \frac{12}{5}$. So $P = (\frac{4}{5}, -\frac{1}{5}, \frac{12}{5})$. Then $a+b+c = \frac{4-1+12}{5} = \frac{15}{5} = 3$.

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a+b+c=$

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