If the points A(-1,0,7), B(3,2, t), C(5, k,-2 | vedclass

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(B) Given that the points $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear,the direction ratios of $AB$ and $BC$ must be proportional.$\frac{3-(-

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$-1: 2$

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(B) Given that the points $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear,the direction ratios of $AB$ and $BC$ must be proportional.$\frac{3-(-1)}{5-3} = \frac{2-0}{k-2} = \frac{t-7}{-2-t}$$\frac{4}{2} = \frac{2}{k-2} = \frac{t-7}{-2-t}$$2 = \frac{2}{k-2} \Rightarrow k-2 = 1 \Rightarrow k = 3$$2 = \frac{t-7}{-2-t} \Rightarrow -4-2t = t-7 \Rightarrow 3t = 3 \Rightarrow t = 1$Thus,the points are $B(3,2,1)$ and $C(5,3,-2)$.The point $P$ is given by $P(t, k-2t, t+k) = P(1, 3-2(1), 1+3) = P(1, 1, 4)$.Let $P$ divide the line segment $BC$ in the ratio $\lambda : 1$.Using the section formula for the $x$-coordinate:$1 = \frac{\lambda(5) + 1(3)}{\lambda + 1}$$\lambda + 1 = 5\lambda + 3$$-2 = 4\lambda$$\lambda = -\frac{1}{2}$Therefore,the required ratio is $-1: 2$.

If the points $A(-1,0,7), B(3,2, t), C(5, k,-2)$ are collinear,then the ratio in which the point $P(t, k-2t, t+k)$ divides the line segment $BC$ is

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