Point and Distance formula Questions in English

(C) The distance of a point $P(x, y, z)$ from the $y$-axis is given by the formula $\sqrt{x^2 + z^2}$.
Substituting the coordinates of the point $(1, 2, 3)$ into the formula:
Distance $= \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
Thus,the distance of the point $(1, 2, 3)$ from the $y$-axis is $\sqrt{10}$.

(C) The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
Given points are $(1, 3, 2)$ and $(2, 1, 3)$.
Substituting the values:
$d = \sqrt{(2 - 1)^2 + (1 - 3)^2 + (3 - 2)^2}$
$d = \sqrt{(1)^2 + (-2)^2 + (1)^2}$
$d = \sqrt{1 + 4 + 1}$
$d = \sqrt{6}$
Thus,the correct option is $C$.

(B) The distance of a point $P(x, y, z)$ from the coordinate axes is given by:
$1$. Distance from the $x$-axis: $d_x = \sqrt{y^2 + z^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
$2$. Distance from the $y$-axis: $d_y = \sqrt{x^2 + z^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
$3$. Distance from the $z$-axis: $d_z = \sqrt{x^2 + y^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.
Thus,the distances are $\sqrt{13}, \sqrt{10}, \sqrt{5}$.

(B) Let the point be $P(x, y, z)$.
The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$.
The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$.
The distance of point $P(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$.
According to the problem,the sum of the squares of these distances is $36$:
$(\sqrt{y^2 + z^2})^2 + (\sqrt{x^2 + z^2})^2 + (\sqrt{x^2 + y^2})^2 = 36$
$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 36$
$2(x^2 + y^2 + z^2) = 36$
$x^2 + y^2 + z^2 = 18$
The distance of the point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $\sqrt{x^2 + y^2 + z^2}$.
Therefore,the distance is $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.

(A) Let the point be $P(x, y, z)$. Since $P$ is equidistant from the given points $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$,and $C(0, 0, c)$,we have $PO^2 = PA^2 = PB^2 = PC^2$.
$x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2 = x^2 + y^2 + (z - c)^2$.
From $x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2$,we get $x^2 = x^2 - 2ax + a^2$,which implies $2ax = a^2$,so $x = \frac{a}{2}$.
Similarly,from $x^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2$,we get $y = \frac{b}{2}$.
And from $x^2 + y^2 + z^2 = x^2 + y^2 + (z - c)^2$,we get $z = \frac{c}{2}$.
Thus,the coordinates of the point are $\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)$.

(A) The coordinates of the given point are $(x, y, z) = (3, 4, 5)$.
The perpendicular distance of a point $(x, y, z)$ from the $y$-axis is given by the formula $d = \sqrt{x^2 + z^2}$.
Substituting the values $x = 3$ and $z = 5$ into the formula:
$d = \sqrt{3^2 + 5^2}$
$d = \sqrt{9 + 25}$
$d = \sqrt{34}$.

(B) When a point moves parallel to the $x$-axis,its distance from the $yz$-plane remains constant.
This means that the coordinates $y$ and $z$ do not change as the point moves.
Only the $x$-coordinate changes as the point moves along the line parallel to the $x$-axis.
Therefore,the variables $y$ and $z$ remain fixed.

(C) The distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is given by the formula:
$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
Given points are $A(1, 2, 3)$ and $B(-1, -1, -1)$.
Substituting the values into the formula:
$AB = \sqrt{(-1 - 1)^2 + (-1 - 2)^2 + (-1 - 3)^2}$
$AB = \sqrt{(-2)^2 + (-3)^2 + (-4)^2}$
$AB = \sqrt{4 + 9 + 16}$
$AB = \sqrt{29}$
Thus,the correct option is $C$.

(C) The distance of a point $P(x, y, z)$ from the $y$-axis is given by the formula $d = \sqrt{x^2 + z^2}$.
Given the point $(4, 3, 5)$,we have $x = 4$,$y = 3$,and $z = 5$.
Substituting these values into the formula:
$d = \sqrt{4^2 + 5^2}$
$d = \sqrt{16 + 25}$
$d = \sqrt{41}$
Thus,the distance is $\sqrt{41}$.

(D) Given points are $A(a, 2, 1), B(1, -1, 1), C(2, -3, 4), D(a+1, a+2, a+3)$.
First,calculate $AB = 5$:
$AB = \sqrt{(a-1)^2 + (2 - (-1))^2 + (1-1)^2} = 5$
$\sqrt{(a-1)^2 + 3^2 + 0} = 5$
$(a-1)^2 + 9 = 25$
$(a-1)^2 = 16$
$a-1 = \pm 4$
$a = 5$ or $a = -3$ $(i)$
Next,calculate $CD = 6$:
$CD = \sqrt{(a+1-2)^2 + (a+2 - (-3))^2 + (a+3-4)^2} = 6$
$\sqrt{(a-1)^2 + (a+5)^2 + (a-1)^2} = 6$
$(a-1)^2 + (a+5)^2 + (a-1)^2 = 36$
$(a^2 - 2a + 1) + (a^2 + 10a + 25) + (a^2 - 2a + 1) = 36$
$3a^2 + 6a + 27 = 36$
$3a^2 + 6a - 9 = 0$
$a^2 + 2a - 3 = 0$
$(a+3)(a-1) = 0$
$a = -3$ or $a = 1$ $(ii)$
The common solution for both $(i)$ and $(ii)$ is $a = -3$.

(A) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4} \right)$.
Given vertices are $O(0, 0, 0)$,$A(a, 2, 3)$,$B(1, b, 2)$,and $C(2, 1, c)$.
The centroid is $(1, 2, -1)$.
Equating the coordinates:
$\frac{a+1+2+0}{4} = 1 \Rightarrow a+3 = 4 \Rightarrow a = 1$.
$\frac{2+b+1+0}{4} = 2 \Rightarrow b+3 = 8 \Rightarrow b = 5$.
$\frac{3+2+c+0}{4} = -1 \Rightarrow c+5 = -4 \Rightarrow c = -9$.
Thus,point $P$ is $(1, 5, -9)$.
The distance of $P$ from the origin $(0, 0, 0)$ is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + 5^2 + (-9)^2} = \sqrt{1 + 25 + 81} = \sqrt{107}$.

(C) The distance of a point $(x, y, z)$ from the origin $(0, 0, 0)$ is $\sqrt{x^2 + y^2 + z^2}$. For $(1, 2, 3)$,this is $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The distance of a point $(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$. For $(1, 2, 3)$,this is $\sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
The distance of a point $(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$. For $(1, 2, 3)$,this is $\sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
The distance of a point $(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$. For $(1, 2, 3)$,this is $\sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.
Thus,the point $(1, 2, 3)$ is at a distance of $\sqrt{10}$ from the $y$-axis.

(A) In a three-dimensional Cartesian coordinate system,the position of a point $P$ is given by its coordinates $(x, y, z)$.
The $yz$-plane is defined by the equation $x = 0$.
The perpendicular distance of any point $(x, y, z)$ from the plane $ax + by + cz + d = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
For the $yz$-plane,the equation is $1x + 0y + 0z + 0 = 0$.
Substituting the coordinates of point $P(x, y, z)$ into the distance formula:
$d = \frac{|1(x) + 0(y) + 0(z) + 0|}{\sqrt{1^2 + 0^2 + 0^2}}$
$d = \frac{|x|}{1} = |x|$.
Thus,the distance of the point $P(x, y, z)$ from the $yz$-plane is $|x|$.

(A) The distance of a point $P(x_1, y_1, z_1)$ from the $x$-axis is given by the formula $d = \sqrt{y_1^2 + z_1^2}$.
Given the point $P(1, 2, 3)$,we have $x_1 = 1$,$y_1 = 2$,and $z_1 = 3$.
Substituting these values into the formula:
$d = \sqrt{2^2 + 3^2}$
$d = \sqrt{4 + 9}$
$d = \sqrt{13}$ units.
Therefore,the correct option is $A$.

(C) Let the point on the $X$-axis be $P\ (x, 0, 0)$.
Since $P$ is equidistant from $A\ (2, -5, 7)$ and $B\ (1, 3, 6)$,we have $PA = PB$,which implies $PA^2 = PB^2$.
Using the distance formula: $(x - 2)^2 + (0 - (-5))^2 + (0 - 7)^2 = (x - 1)^2 + (0 - 3)^2 + (0 - 6)^2$.
$(x - 2)^2 + 25 + 49 = (x - 1)^2 + 9 + 36$.
$x^2 - 4x + 4 + 74 = x^2 - 2x + 1 + 45$.
$x^2 - 4x + 78 = x^2 - 2x + 46$.
$-4x + 2x = 46 - 78$.
$-2x = -32$.
$x = 16$.
Thus,the point is $P\ (16, 0, 0)$.

(A) Let the coordinates of point $P$ be $(x, y, z)$.
Given $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.
The square of the distance $PA$ is $PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 - 10z + 25 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 50$.
The square of the distance $PB$ is $PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 + z^2 + 14z + 49 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 59$.
Substituting these into the equation $PA^2 - PB^2 + 2k^2 = 0$:
$(x^2 + y^2 + z^2 - 6x - 8y - 10z + 50) - (x^2 + y^2 + z^2 + 2x - 6y + 14z + 59) + 2k^2 = 0$.
Simplifying the expression:
$(-6x - 2x) + (-8y + 6y) + (-10z - 14z) + (50 - 59) + 2k^2 = 0$.
$-8x - 2y - 24z - 9 + 2k^2 = 0$.
Rearranging the terms:
$8x + 2y + 24z = 2k^2 - 9$.

(C) Let the coordinates of the point be $P(x, y, z)$.
The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$. The square of this distance is $y^2 + z^2$.
The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$. The square of this distance is $x^2 + z^2$.
The distance of point $P(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$. The square of this distance is $x^2 + y^2$.
According to the problem,the sum of the squares of these distances is $36$:
$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 36$
Simplifying the equation:
$2x^2 + 2y^2 + 2z^2 = 36$
$x^2 + y^2 + z^2 = 18$
The distance of the point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $d = \sqrt{x^2 + y^2 + z^2}$.
Substituting the value:
$d = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.

(A) The distance of a point $P(a, b, c)$ from the $z$-axis is the distance between $P(a, b, c)$ and the foot of the perpendicular from $P$ onto the $z$-axis.
The coordinates of the foot of the perpendicular from $P(a, b, c)$ onto the $z$-axis are $(0, 0, c)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$,we get:
$d = \sqrt{(a - 0)^2 + (b - 0)^2 + (c - c)^2}$
$d = \sqrt{a^2 + b^2 + 0^2}$
$d = \sqrt{a^2 + b^2}$
Thus,the distance is $\sqrt{a^2 + b^2}$.

(B) Let the coordinates of the point be $P(x, y, z)$.
Since the point lies in the $yz$-plane,its $x$-coordinate must be $0$. So,$P = (0, y, z)$.
The sum of the coordinates is $y + z = 3$ (Equation $1$).
The distance of point $(0, y, z)$ from the $xz$-plane is $|y|$.
The distance of point $(0, y, z)$ from the $xy$-plane is $|z|$.
According to the problem,the distance from the $xz$-plane is twice the distance from the $xy$-plane,so $|y| = 2|z|$.
Case $1$: $y = 2z$.
Substituting into Equation $1$: $2z + z = 3 \implies 3z = 3 \implies z = 1$.
Then $y = 2(1) = 2$.
The point is $(0, 2, 1)$.
Case $2$: $y = -2z$.
Substituting into Equation $1$: $-2z + z = 3 \implies -z = 3 \implies z = -3$.
Then $y = -2(-3) = 6$.
The point is $(0, 6, -3)$.
Comparing with the given options,the point $(0, 2, 1)$ is present.

(A) Let the coordinates of point $P$ be $(x, y, z)$ such that $PA = PB = PC = PO$.
From $PO^2 = PA^2$,we have:
$x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2$
$x^2 = x^2 - 2ax + a^2$
$2ax = a^2 \implies x = \frac{a}{2}$ (since $a \neq 0$).
Similarly,from $PO^2 = PB^2$,we have:
$x^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2$
$y^2 = y^2 - 2by + b^2$
$2by = b^2 \implies y = \frac{b}{2}$ (since $b \neq 0$).
Similarly,from $PO^2 = PC^2$,we have:
$x^2 + y^2 + z^2 = x^2 + y^2 + (z - c)^2$
$z^2 = z^2 - 2cz + c^2$
$2cz = c^2 \implies z = \frac{c}{2}$ (since $c \neq 0$).
Therefore,the coordinates of point $P$ are $\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)$.

(A) Given the points $P(5, 4, a)$ and $Q(-1, 2, -2)$ with distance $PQ = 7$.
Using the distance formula: $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.
Substituting the values: $7^2 = (-1 - 5)^2 + (2 - 4)^2 + (-2 - a)^2$.
$49 = (-6)^2 + (-2)^2 + (-(2 + a))^2$.
$49 = 36 + 4 + (a + 2)^2$.
$49 = 40 + (a + 2)^2$.
$(a + 2)^2 = 49 - 40 = 9$.
Taking the square root on both sides: $a + 2 = \pm 3$.
Case $1$: $a + 2 = 3 \implies a = 1$.
Case $2$: $a + 2 = -3 \implies a = -5$.
Therefore,the values of $a$ are $-5$ and $1$.

(B) The coordinates of the point are $P(a, b, c)$.
To find the shortest distance from a point $P(a, b, c)$ to the $X-$axis,we consider the projection of the point onto the $X-$axis.
The projection of point $P(a, b, c)$ on the $X-$axis is $P'(a, 0, 0)$.
The distance $d$ between $P(a, b, c)$ and $P'(a, 0, 0)$ is given by the distance formula:
$d = \sqrt{(a - a)^2 + (b - 0)^2 + (c - 0)^2}$
$d = \sqrt{0^2 + b^2 + c^2}$
$d = \sqrt{b^2 + c^2}$
Therefore,the shortest distance is $\sqrt{b^2 + c^2}$.

(A) The point $P$ has coordinates $(x, y, z) = (2, 4, 5)$.
This means the distance along the $X$-axis is $2$,along the $Y$-axis is $4$,and along the $Z$-axis is $5$.
The point $F$ lies on the $XZ$-plane,which means its $Y$-coordinate is $0$.
Since $F$ is directly above the $X$-axis at the same height as $P$ and at the same distance along the $X$-axis as $P$,its coordinates are $(2, 0, 5)$.

(A) In a three-dimensional Cartesian coordinate system,any point lying on the $x$-axis has its distance from both the $y$-axis and $z$-axis equal to $0$.
Therefore,for any point on the $x$-axis,the coordinates are of the form $(x, 0, 0)$.
Thus,the $y$-coordinate is $0$ and the $z$-coordinate is $0$.

(A) The $x$-axis and $y$-axis taken together determine a plane known as the $XY$-plane.

(A) The distance $PQ$ between the points $P(x_1, y_1, z_1) = (1, -3, 4)$ and $Q(x_2, y_2, z_2) = (-4, 1, 2)$ is given by the distance formula:
$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
$PQ = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}$
$PQ = \sqrt{(-5)^2 + (4)^2 + (-2)^2}$
$PQ = \sqrt{25 + 16 + 4}$
$PQ = \sqrt{45}$
$PQ = 3\sqrt{5} \text{ units}$

Points are collinear if they lie on the same straight line.
First,we calculate the distances between the points using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
$QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$.
$PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$.
Since $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$,the sum of the lengths of two segments equals the length of the third segment.
Therefore,the points $P$,$Q$,and $R$ are collinear.

(A) Let the coordinates of point $P$ be $(x, y, z)$.
$PA^{2} = (x-3)^{2} + (y-4)^{2} + (z-5)^{2} = x^{2} - 6x + 9 + y^{2} - 8y + 16 + z^{2} - 10z + 25 = x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50$.
$PB^{2} = (x+1)^{2} + (y-3)^{2} + (z+7)^{2} = x^{2} + 2x + 1 + y^{2} - 6y + 9 + z^{2} + 14z + 49 = x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59$.
Given $PA^{2} + PB^{2} = 2k^{2}$,we have:
$(x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50) + (x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59) = 2k^{2}$.
Combining like terms:
$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z + 109 = 2k^{2}$.
Thus,$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z = 2k^{2} - 109$.

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:
$PQ = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$
Given points are $(2, 3, 5)$ and $(4, 3, 1)$.
Substituting the values into the formula:
$PQ = \sqrt{(4-2)^{2} + (3-3)^{2} + (1-5)^{2}}$
$PQ = \sqrt{(2)^{2} + (0)^{2} + (-4)^{2}}$
$PQ = \sqrt{4 + 0 + 16}$
$PQ = \sqrt{20}$
$PQ = 2 \sqrt{5}$

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:
$PQ = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$
Given points are $(-3, 7, 2)$ and $(2, 4, -1)$.
Here,$x_{1} = -3, y_{1} = 7, z_{1} = 2$ and $x_{2} = 2, y_{2} = 4, z_{2} = -1$.
Substituting these values into the distance formula:
$PQ = \sqrt{(2 - (-3))^{2} + (4 - 7)^{2} + (-1 - 2)^{2}}$
$PQ = \sqrt{(2 + 3)^{2} + (-3)^{2} + (-3)^{2}}$
$PQ = \sqrt{(5)^{2} + 9 + 9}$
$PQ = \sqrt{25 + 9 + 9}$
$PQ = \sqrt{43}$

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:
$d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$
Given points are $P(-1, 3, -4)$ and $Q(1, -3, 4)$.
Substituting the values into the formula:
$d = \sqrt{(1 - (-1))^{2} + (-3 - 3)^{2} + (4 - (-4))^{2}}$
$d = \sqrt{(1 + 1)^{2} + (-6)^{2} + (4 + 4)^{2}}$
$d = \sqrt{(2)^{2} + (-6)^{2} + (8)^{2}}$
$d = \sqrt{4 + 36 + 64}$
$d = \sqrt{104}$
$d = \sqrt{4 \times 26} = 2 \sqrt{26}$

(A) The distance between points $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ is given by the formula:
$PQ = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$
Given points are $(2, -1, 3)$ and $(-2, 1, 3)$.
Substituting the values into the formula:
$Distance = \sqrt{(-2 - 2)^{2} + (1 - (-1))^{2} + (3 - 3)^{2}}$
$= \sqrt{(-4)^{2} + (2)^{2} + (0)^{2}}$
$= \sqrt{16 + 4 + 0}$
$= \sqrt{20}$
$= \sqrt{4 \times 5} = 2 \sqrt{5}$

Let the points $(-2, 3, 5), (1, 2, 3),$ and $(7, 0, -1)$ be denoted by $P, Q,$ and $R$ respectively.
Points $P, Q,$ and $R$ are collinear if they lie on the same straight line.
$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$
$QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$
$PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
Since $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$,the sum of the distances between two pairs of points is equal to the distance between the third pair.
Therefore,the points $P, Q,$ and $R$ are collinear.

(A) If a point lies on the $y$-axis,its $x$-coordinate and $z$-coordinate must be $0$.
Let the point be $A(0, b, 0)$.
The distance between $A(0, b, 0)$ and $P(3, -2, 5)$ is given as $5\sqrt{2}$.
Using the distance formula,$AP = \sqrt{(3-0)^2 + (-2-b)^2 + (5-0)^2} = 5\sqrt{2}$.
Squaring both sides,we get $(3)^2 + (-2-b)^2 + (5)^2 = (5\sqrt{2})^2$.
$9 + (4 + 4b + b^2) + 25 = 50$.
$b^2 + 4b + 38 = 50$.
$b^2 + 4b - 12 = 0$.
Factoring the quadratic equation: $(b + 6)(b - 2) = 0$.
Thus,$b = 2$ or $b = -6$.
The coordinates of the points are $(0, 2, 0)$ and $(0, -6, 0)$.

(N/A) To locate a point $(x, y, z)$ in a three-dimensional space:
$1.$ Start at the origin $(0, 0, 0)$.
$2.$ Move $x$ units along the $x$-axis (positive or negative).
$3.$ From that position,move $y$ units parallel to the $y$-axis.
$4.$ Finally,move $z$ units parallel to the $z$-axis.
The points are plotted as follows:
- Point $A(1, -1, 3)$ lies in the octant where $x > 0, y < 0, z > 0$.
- Point $B(-1, 2, 4)$ lies in the octant where $x < 0, y > 0, z > 0$.
- Point $C(-2, -4, -7)$ lies in the octant where $x < 0, y < 0, z < 0$.
- Point $D(-4, 2, -5)$ lies in the octant where $x < 0, y > 0, z < 0$.

(N/A) The octants are determined by the signs of the coordinates $(x, y, z)$ as follows:

Octant	Signs $(x, y, z)$
$I$	$(+, +, +)$
$II$	$(-, +, +)$
$III$	$(-, -, +)$
$IV$	$(+, -, +)$
$V$	$(+, +, -)$
$VI$	$(-, +, -)$
$VII$	$(-, -, -)$
$VIII$	$(+, -, -)$

Based on the signs of the coordinates:
$(i) (1, 2, 3)$ lies in $I$ octant.
$(ii) (4, -2, 3)$ lies in $IV$ octant.
$(iii) (4, -2, -5)$ lies in $VIII$ octant.
$(iv) (4, 2, -5)$ lies in $V$ octant.
$(v) (-4, 2, 5)$ lies in $II$ octant.
$(vi) (-3, -1, 6)$ lies in $III$ octant.
$(vii) (2, -4, -7)$ lies in $VIII$ octant.
$(viii) (-4, 2, -5)$ lies in $VI$ octant.

(N/A) The coordinates of the feet of perpendiculars from a point $P(x, y, z)$ onto the axes are:
- On the $x$-axis,the foot of the perpendicular is $A(x, 0, 0)$.
- On the $y$-axis,the foot of the perpendicular is $B(0, y, 0)$.
- On the $z$-axis,the foot of the perpendicular is $C(0, 0, z)$.
Applying this to the given points:
$(i)$ For $P(3, 4, 2)$: $A(3, 0, 0), B(0, 4, 0), C(0, 0, 2)$.
$(ii)$ For $P(-5, 3, 7)$: $A(-5, 0, 0), B(0, 3, 0), C(0, 0, 7)$.
$(iii)$ For $P(4, -3, -5)$: $A(4, 0, 0), B(0, -3, 0), C(0, 0, -5)$.

(N/A) We know that on the $xy$-plane,$z=0$; on the $yz$-plane,$x=0$; and on the $zx$-plane,$y=0$.
Thus,the coordinates of the feet of the perpendiculars from point $P(x, y, z)$ are $A(x, y, 0)$ on the $xy$-plane,$B(0, y, z)$ on the $yz$-plane,and $C(x, 0, z)$ on the $zx$-plane.
$(i)$ For $P(3, 4, 5)$:
$A(3, 4, 0), B(0, 4, 5), C(3, 0, 5)$
(ii) For $P(-5, 3, 7)$:
$A(-5, 3, 0), B(0, 3, 7), C(-5, 0, 7)$
(iii) For $P(4, -3, -5)$:
$A(4, -3, 0), B(0, -3, -5), C(4, 0, -5)$

(C) Let the points be $A = (2, 0, 0)$ and $B = (-3, 0, 0)$.
Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in $3D$ space:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
Substituting the values:
$d = \sqrt{(-3 - 2)^2 + (0 - 0)^2 + (0 - 0)^2}$
$d = \sqrt{(-5)^2 + 0 + 0}$
$d = \sqrt{25} = 5$
Thus,the distance between the points is $5$ units.

(B) The distance $d$ of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$ is given by the formula:
$d = \sqrt{x^2 + y^2 + z^2}$
Given the point $(6, 6, 7)$,we substitute the values into the formula:
$d = \sqrt{6^2 + 6^2 + 7^2}$
$d = \sqrt{36 + 36 + 49}$
$d = \sqrt{121}$
$d = 11$
Thus,the distance from the origin to the point $(6, 6, 7)$ is $11$ units.

(B) The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$.
The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$.
The distance of point $P(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$.
According to the problem,the sum of the squares of these distances is $242$:
$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 242$
$2(x^2 + y^2 + z^2) = 242$
$x^2 + y^2 + z^2 = 121$
The distance of point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $d = \sqrt{x^2 + y^2 + z^2}$.
Substituting the value,$d = \sqrt{121} = 11$.
Thus,the distance is $11$ units.

(C) The distance of a point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$.
The distance of point $P$ from the $y$-axis is $\sqrt{x^2 + z^2}$.
The distance of point $P$ from the $z$-axis is $\sqrt{x^2 + y^2}$.
Given that the sum of the squares of these distances is $324$:
$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 324$
$2(x^2 + y^2 + z^2) = 324$
$x^2 + y^2 + z^2 = 162$
The distance of point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $d = \sqrt{x^2 + y^2 + z^2}$.
Substituting the value,we get $d = \sqrt{162} = \sqrt{81 \times 2} = 9 \sqrt{2}$.

(A) The coordinates of the point $P$ are $(a, b, c)$.
To find the distance of point $P$ from the $x$-axis,we project the point onto the $x$-axis.
The projection of point $P(a, b, c)$ on the $x$-axis is the point $A(a, 0, 0)$.
The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
Substituting the coordinates of $P(a, b, c)$ and $A(a, 0, 0)$:
$d = \sqrt{(a-a)^2 + (0-b)^2 + (0-c)^2}$
$d = \sqrt{0^2 + (-b)^2 + (-c)^2}$
$d = \sqrt{b^2 + c^2}$.

(D) The distance of a point $P(x, y, z)$ from the $yz$-plane is given by the absolute value of its $x$-coordinate,which is $|x|$.
Given the point $P(-3, 4, 5)$,the $x$-coordinate is $-3$.
Therefore,the distance from the $yz$-plane is $|-3| = 3 \text{ units}$.
Thus,the correct option is $D$.

(C) In a three-dimensional Cartesian coordinate system,the octants are determined by the signs of the coordinates $(x, y, z)$.
For the point $(1, -3, 4)$,we have $x > 0$,$y < 0$,and $z > 0$.
The octants are defined as follows:
$I: (+, +, +)$
$II: (-, +, +)$
$III: (-, -, +)$
$IV: (+, -, +)$
$V: (+, +, -)$
$VI: (-, +, -)$
$VII: (-, -, -)$
$VIII: (+, -, -)$
Since the signs are $(+, -, +)$,the point $(1, -3, 4)$ lies in the $IV$ octant.

(B) The distance of a point $(x, y, z)$ from the $X, Y, Z$-axes are given by $d_1 = \sqrt{y^2 + z^2}$,$d_2 = \sqrt{x^2 + z^2}$,and $d_3 = \sqrt{x^2 + y^2}$.
For the point $(1, 2, 3)$:
$d_1 = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \implies d_1^2 = 13$.
$d_2 = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \implies d_2^2 = 10$.
$d_3 = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \implies d_3^2 = 5$.
Now,calculate $2 d_2^2 + d_3^2 + 1$:
$2(10) + 5 + 1 = 20 + 5 + 1 = 26$.
Since $d_1^2 = 13$,we have $26 = 2 \times 13 = 2 d_1^2$.

(B) Let the side of the square be $a$.
The length of the diagonal $d$ of a square with side $a$ is given by $d = a\sqrt{2}$.
The distance between the two given points $(1, -2, 3)$ and $(2, -3, 5)$ is the length of the diagonal $d$.
$d = \sqrt{(2-1)^2 + (-3 - (-2))^2 + (5-3)^2}$
$d = \sqrt{(1)^2 + (-1)^2 + (2)^2}$
$d = \sqrt{1 + 1 + 4} = \sqrt{6}$
Since $d = a\sqrt{2}$,we have $a\sqrt{2} = \sqrt{6}$.
Dividing both sides by $\sqrt{2}$,we get $a = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$.

(A) Let the point be $P(x, y, z)$.
$(a)$ The perpendicular distance $d_1$ of point $P$ from the $x$-axis is $\sqrt{y^2 + z^2}$. So,$d_1^2 = y^2 + z^2$.
$(b)$ The perpendicular distance $d_2$ of point $P$ from the $y$-axis is $\sqrt{x^2 + z^2}$. So,$d_2^2 = x^2 + z^2$.
$(c)$ The perpendicular distance $d_3$ of point $P$ from the $z$-axis is $\sqrt{x^2 + y^2}$. So,$d_3^2 = x^2 + y^2$.
The distance $d$ of point $P$ from the origin $(0, 0, 0)$ is $\sqrt{x^2 + y^2 + z^2}$. So,$d^2 = x^2 + y^2 + z^2$.
The sum of the squares of the perpendicular distances is $d_1^2 + d_2^2 + d_3^2 = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2(x^2 + y^2 + z^2)$.
Given that this sum is $k$ times the square of the distance from the origin,we have $2(x^2 + y^2 + z^2) = k(x^2 + y^2 + z^2)$.
Therefore,$k = 2$.