Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a', b', c'$ from the origin along the axes,then:

  • A
    $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2} = 0$
  • B
    $\frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{c^2} + \frac{1}{a'^2} + \frac{1}{b'^2} - \frac{1}{c'^2} = 0$
  • C
    $\frac{1}{a^2} - \frac{1}{b^2} - \frac{1}{c^2} + \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$
  • D
    $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$

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$A$ variable plane at a constant distance $p$ from the origin meets the coordinate axes at points $A, B, C$. Through these points,planes are drawn parallel to the coordinate planes. Find the locus of their point of intersection.

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Let $R^3$ denote the three-dimensional space. Take two points $P=(1, 2, 3)$ and $Q=(4, 2, 7)$. Let $\operatorname{dist}(X, Y)$ denote the distance between two points $X$ and $Y$ in $R^3$. Let
$S=\{X \in R^3: (\operatorname{dist}(X, P))^2 - (\operatorname{dist}(X, Q))^2 = 50\}$
$T=\{Y \in R^3: (\operatorname{dist}(Y, Q))^2 - (\operatorname{dist}(Y, P))^2 = 50\}$
Then which of the following statements is (are) $TRUE$?
$(A)$ There is a triangle whose area is $1$ and all of whose vertices are from $S$.
$(B)$ There are two distinct points $L$ and $M$ in $T$ such that each point on the line segment $LM$ is also in $T$.
$(C)$ There are infinitely many rectangles of perimeter $48$,two of whose vertices are from $S$ and the other two vertices are from $T$.
$(D)$ There is a square of perimeter $48$,two of whose vertices are from $S$ and the other two vertices are from $T$.

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

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The distance of the plane $6x - 3y + 2z - 14 = 0$ from the origin is

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