If the line joining (2,3,-1) and (3,5,-3) is | vedclass

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(D) Let the points be $P(2,3,-1)$ and $Q(3,5,-3)$. The direction ratios of the line $PQ$ are $(3-2, 5-3, -3-(-1)) = (1, 2, -2)$.Let the points be $A(1

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$(3,5,7)$

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(D) Let the points be $P(2,3,-1)$ and $Q(3,5,-3)$. The direction ratios of the line $PQ$ are $(3-2, 5-3, -3-(-1)) = (1, 2, -2)$.Let the points be $A(1,2,3)$ and $B(\alpha, \beta, \gamma)$. The direction ratios of the line $AB$ are $(\alpha-1, \beta-2, \gamma-3)$.Since the lines are perpendicular,the dot product of their direction ratios must be zero:$1(\alpha-1) + 2(\beta-2) - 2(\gamma-3) = 0$$\alpha - 1 + 2\beta - 4 - 2\gamma + 6 = 0$$\alpha + 2\beta - 2\gamma + 1 = 0$.Now,we check the options for point $B(\alpha, \beta, \gamma)$:For option $A(-3,5,7)$: $-3 + 2(5) - 2(7) + 1 = -3 + 10 - 14 + 1 = -6 
eq 0$.For option $B(3,-5,7)$: $3 + 2(-5) - 2(7) + 1 = 3 - 10 - 14 + 1 = -20 
eq 0$.For option $C(3,5,-7)$: $3 + 2(5) - 2(-7) + 1 = 3 + 10 + 14 + 1 = 28 
eq 0$.For option $D(3,5,7)$: $3 + 2(5) - 2(7) + 1 = 3 + 10 - 14 + 1 = 0$.Since the condition is satisfied for option $D$,the correct point is $(3,5,7)$.

If the line joining $(2,3,-1)$ and $(3,5,-3)$ is perpendicular to the line joining $A(1,2,3)$ and $B(\alpha, \beta, \gamma)$,then a possible point for $B$ is

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