Statement-1: The point A(1, 0, 7) is the refl | vedclass

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(D) Let the line be $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = k$. Any point on the line is $P(k, 2k+1, 3k+2)$.For $P$ to be the midpoint of $A

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Statement-$1$ is true,Statement-$2$ is true,Statement-$2$ is a correct explanation for Statement-$1$.

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(D) Let the line be $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = k$. Any point on the line is $P(k, 2k+1, 3k+2)$.For $P$ to be the midpoint of $AB$,where $A(1, 0, 7)$ and $B(1, 6, 3)$,the midpoint $M$ is $(\frac{1+1}{2}, \frac{0+6}{2}, \frac{7+3}{2}) = (1, 3, 5)$.Setting $P = M$,we get $k=1, 2k+1=3 \Rightarrow k=1, 3k+2=5 \Rightarrow k=1$. Thus,$M$ lies on the line.The direction vector of $AB$ is $\vec{AB} = (1-1, 6-0, 3-7) = (0, 6, -4)$.The direction vector of the line is $\vec{v} = (1, 2, 3)$.For the line to be the perpendicular bisector,$\vec{AB} \cdot \vec{v} = 0$. $(0)(1) + (6)(2) + (-4)(3) = 0 + 12 - 12 = 0$. Since the midpoint lies on the line and the line is perpendicular to $AB$,Statement-$2$ is true.Since Statement-$2$ is true,$A$ is the reflection of $B$ in the line,so Statement-$1$ is true and Statement-$2$ is the correct explanation.

Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

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Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.