If A(-2, 4, a),B(1, b, 3),C(c, 0, 4),and D(-5 | vedclass

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(B) Since the points $A, B, C, D$ are collinear,they lie on the same line passing through $D(-5, 6, 1)$.Let the direction ratios of the line be $(p, q

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$8$

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(B) Since the points $A, B, C, D$ are collinear,they lie on the same line passing through $D(-5, 6, 1)$.Let the direction ratios of the line be $(p, q, r)$. The equation of the line is $\frac{x+5}{p} = \frac{y-6}{q} = \frac{z-1}{r} = k$.For point $A(-2, 4, a)$: $\frac{-2+5}{p} = \frac{4-6}{q} = \frac{a-1}{r} \Rightarrow \frac{3}{p} = \frac{-2}{q} = \frac{a-1}{r}$.For point $B(1, b, 3)$: $\frac{1+5}{p} = \frac{b-6}{q} = \frac{3-1}{r} \Rightarrow \frac{6}{p} = \frac{b-6}{q} = \frac{2}{r}$.For point $C(c, 0, 4)$: $\frac{c+5}{p} = \frac{0-6}{q} = \frac{4-1}{r} \Rightarrow \frac{c+5}{p} = \frac{-6}{q} = \frac{3}{r}$.From $\frac{3}{p} = \frac{-2}{q}$ and $\frac{c+5}{p} = \frac{-6}{q}$,we have $\frac{c+5}{3} = \frac{-6}{-2} = 3$ $\Rightarrow c+5 = 9$ $\Rightarrow c = 4$.From $\frac{6}{p} = \frac{2}{r}$ and $\frac{3}{p} = \frac{a-1}{r}$,we have $\frac{a-1}{3} = \frac{2}{6} = \frac{1}{3}$ $\Rightarrow a-1 = 1$ $\Rightarrow a = 2$.From $\frac{6}{p} = \frac{b-6}{q}$ and $\frac{3}{p} = \frac{-2}{q}$,we have $\frac{b-6}{3} = \frac{-2}{-2} = 1$ $\Rightarrow b-6 = 3$ $\Rightarrow b = 9$ (Wait,re-evaluating: $\frac{b-6}{q} = \frac{6}{p}$ and $\frac{-2}{q} = \frac{3}{p}$ $\Rightarrow \frac{b-6}{-2} = \frac{6}{3} = 2$ $\Rightarrow b-6 = -4$ $\Rightarrow b = 2$).Thus,$a=2, b=2, c=4$.Therefore,$a+b+c = 2+2+4 = 8$.

If $A(-2, 4, a)$,$B(1, b, 3)$,$C(c, 0, 4)$,and $D(-5, 6, 1)$ are collinear points,then $a+b+c=$

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