Sphere Questions in English

(D) Let $P$ be any point on the sphere with position vector $r$. Let $A$ and $B$ be the extremities of a diameter with position vectors $a$ and $b$ respectively.
Since $AB$ is a diameter,the angle subtended by the diameter at any point $P$ on the sphere is a right angle $(90^{\circ})$.
Therefore,the vector $\vec{AP}$ is perpendicular to the vector $\vec{BP}$.
The vector $\vec{AP} = r - a$ and the vector $\vec{BP} = r - b$.
Since they are perpendicular,their dot product must be zero:
$(r - a) \cdot (r - b) = 0$.

(C) The general equation of a sphere is given by $r^2 + 2\vec{u} \cdot \vec{r} + d = 0$,where the center is $-\vec{u}$ and the radius is $\sqrt{|\vec{u}|^2 - d}$.
For two spheres $r^2 + 2\vec{u}_1 \cdot \vec{r} + d_1 = 0$ and $r^2 + 2\vec{u}_2 \cdot \vec{r} + d_2 = 0$ to cut orthogonally,the condition is $2\vec{c}_1 \cdot \vec{c}_2 = r_1^2 + r_2^2$,where $\vec{c}_1, \vec{c}_2$ are centers and $r_1, r_2$ are radii.
Here,$\vec{c}_1 = -\vec{u}_1$,$\vec{c}_2 = -\vec{u}_2$,$r_1^2 = |\vec{u}_1|^2 - d_1$,and $r_2^2 = |\vec{u}_2|^2 - d_2$.
Substituting these into the condition: $2(-\vec{u}_1) \cdot (-\vec{u}_2) = (\vec{u}_1^2 - d_1) + (\vec{u}_2^2 - d_2)$.
This simplifies to $2\vec{u}_1 \cdot \vec{u}_2 = |\vec{u}_1|^2 + |\vec{u}_2|^2 - (d_1 + d_2)$.
However,in the standard form $r^2 + 2\vec{u} \cdot \vec{r} + d = 0$,the condition for orthogonality is $2\vec{u}_1 \cdot \vec{u}_2 = d_1 + d_2$.

(C) The general equation of a sphere in vector form is given by $|r|^2 - 2(r \cdot a) + c = 0$,where the radius is $\sqrt{|a|^2 - c}$.
Comparing the given equation $|r|^2 - r \cdot (2i + 4j - 2k) - 10 = 0$ with the standard form:
Here,$2a = (2i + 4j - 2k)$,which implies $a = (i + 2j - k)$.
Also,$c = -10$.
The radius of the sphere is $\sqrt{|a|^2 - c}$.
Calculating $|a|^2 = |i + 2j - k|^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$.
Substituting the values,the radius is $\sqrt{6 - (-10)} = \sqrt{6 + 10} = \sqrt{16} = 4$.
Thus,the equation represents a sphere of radius $4$.

(A) The given equation of the sphere is $\alpha \,r^2 - 2u \cdot r = \beta$.
Dividing the entire equation by $\alpha$ (since $\alpha \ne 0$),we get:
$r^2 - \frac{2}{\alpha} u \cdot r = \frac{\beta}{\alpha}$.
Comparing this with the standard vector equation of a sphere $r^2 - 2a \cdot r = c$,where $a$ is the position vector of the centre,we have:
$2a = \frac{2}{\alpha} u$.
Therefore,$a = \frac{u}{\alpha}$.
Thus,the centre of the sphere is $u/\alpha$.

(D) The equation of the sphere is $|r| = 5$,which represents a sphere with centre at the origin $(0, 0, 0)$ and radius $R = 5$.
The given plane is $r \cdot (i + j + k) = 3\sqrt{3}$.
Let $M$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane. The length of the perpendicular $d$ from the origin to the plane $r \cdot n = p$ is given by $d = \frac{|p|}{|n|}$.
Here,$p = 3\sqrt{3}$ and $n = i + j + k$,so $|n| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Thus,$d = OM = \frac{3\sqrt{3}}{\sqrt{3}} = 3$.
Let $r_{c}$ be the radius of the circular section. In the right-angled triangle $\Delta OPM$,where $P$ is a point on the circumference of the circular section,$OP$ is the radius of the sphere $(R = 5)$ and $OM$ is the distance from the centre to the plane $(d = 3)$.
By the Pythagorean theorem,$OP^2 = OM^2 + r_{c}^2$.
$5^2 = 3^2 + r_{c}^2$
$25 = 9 + r_{c}^2$
$r_{c}^2 = 16$
$r_{c} = 4$.
Therefore,the radius of the circular section is $4$.

(D) The general second-degree equation in three variables is given by $ax^2 + by^2 + cz^2 + 2fyz + 2gxz + 2hxy + 2ux + 2vy + 2wz + d = 0$.
For this equation to represent a sphere,the coefficients of the squared terms must be equal,i.e.,$a = b = c \neq 0$.
Additionally,there should be no product terms of the variables (like $xy$,$yz$,or $zx$),which implies that the coefficients of these terms must be zero,i.e.,$f = g = h = 0$.
Therefore,the conditions are $a = b = c$ and $f = g = h = 0$.

(C) Let the point be $P(x, y, z)$. The given condition is $PF_1 + PF_2 = 10$,where $F_1 = (4, 0, 0)$ and $F_2 = (-4, 0, 0)$.
$\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$
Let $S = x^2 + y^2 + z^2 + 16$. Then the equation is $\sqrt{S - 8x} + \sqrt{S + 8x} = 10$.
Squaring both sides: $(S - 8x) + (S + 8x) + 2\sqrt{S^2 - 64x^2} = 100$
$2S + 2\sqrt{S^2 - 64x^2} = 100 \Rightarrow S + \sqrt{S^2 - 64x^2} = 50$
$\sqrt{S^2 - 64x^2} = 50 - S$
Squaring again: $S^2 - 64x^2 = 2500 - 100S + S^2$
$100S - 64x^2 = 2500$
Substitute $S = x^2 + y^2 + z^2 + 16$:
$100(x^2 + y^2 + z^2 + 16) - 64x^2 = 2500$
$100x^2 + 100y^2 + 100z^2 + 1600 - 64x^2 = 2500$
$36x^2 + 100y^2 + 100z^2 = 900$
Dividing by $4$: $9x^2 + 25y^2 + 25z^2 = 225$.

(C) Let the two given points be $A(a, 0, 0)$ and $B(-a, 0, 0)$ in a $3D$ coordinate system. Let the moving point be $P(x, y, z)$.
According to the problem,$PA^2 + PB^2 = k$,where $k$ is a constant.
$(x-a)^2 + y^2 + z^2 + (x+a)^2 + y^2 + z^2 = k$
$(x^2 - 2ax + a^2 + y^2 + z^2) + (x^2 + 2ax + a^2 + y^2 + z^2) = k$
$2x^2 + 2y^2 + 2z^2 + 2a^2 = k$
$x^2 + y^2 + z^2 = \frac{k - 2a^2}{2}$
This equation represents a sphere with its center at the origin $(0, 0, 0)$.

(A) The given equation is ${x^2} + {y^2} + {z^2} = 0$.
Since $x^2$,$y^2$,and $z^2$ are all non-negative real numbers for any real values of $x$,$y$,and $z$,their sum can be zero if and only if each term is individually zero.
Therefore,$x^2 = 0$,$y^2 = 0$,and $z^2 = 0$.
This implies $x = 0$,$y = 0$,and $z = 0$.
Thus,the equation represents the point $(0, 0, 0)$ in three-dimensional space.

(A) Given the equation is ${x^2} + {y^2} + {z^2} + 1 = 0$.
For any real values of $x, y, z$,the squares ${x^2}, {y^2}, {z^2}$ are always non-negative,i.e.,${x^2} \ge 0, {y^2} \ge 0, {z^2} \ge 0$.
Therefore,the sum ${x^2} + {y^2} + {z^2} \ge 0$.
Adding $1$ to both sides,we get ${x^2} + {y^2} + {z^2} + 1 \ge 1$.
Since the expression is always greater than or equal to $1$,it can never be equal to $0$.
Thus,there are no real coordinates $(x, y, z)$ that satisfy this equation.
Hence,the locus is an empty set.

(A) Let the general equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$.
Since the sphere passes through the origin $(0, 0, 0)$,we substitute $x=0, y=0, z=0$ to get $d = 0$.
Since it passes through $(0, 2, 0)$,we have $0^2 + 2^2 + 0^2 + 2u(0) + 2v(2) + 2w(0) + 0 = 0$,which simplifies to $4 + 4v = 0$,so $v = -1$.
Since it passes through $(1, 0, 0)$,we have $1^2 + 0^2 + 0^2 + 2u(1) + 2v(0) + 2w(0) + 0 = 0$,which simplifies to $1 + 2u = 0$,so $u = -1/2$.
Since it passes through $(0, 0, 4)$,we have $0^2 + 0^2 + 4^2 + 2u(0) + 2v(0) + 2w(4) + 0 = 0$,which simplifies to $16 + 8w = 0$,so $w = -2$.
The centre of the sphere is given by $(-u, -v, -w)$.
Substituting the values,the centre is $\left( -(-1/2), -(-1), -(-2) \right) = \left( \frac{1}{2}, 1, 2 \right)$.

(B) sphere touching the three coordinate planes has its center at $(a, a, a)$ and radius $a$ (assuming the sphere is in the first octant).
The standard equation of a sphere with center $(h, k, l)$ and radius $r$ is $(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$.
Substituting the center $(a, a, a)$ and radius $r = a$,we get:
$(x - a)^2 + (y - a)^2 + (z - a)^2 = a^2$
Expanding the squares:
$(x^2 - 2ax + a^2) + (y^2 - 2ay + a^2) + (z^2 - 2az + a^2) = a^2$
Simplifying the equation:
$x^2 + y^2 + z^2 - 2ax - 2ay - 2az + 3a^2 = a^2$
$x^2 + y^2 + z^2 - 2a(x + y + z) + 2a^2 = 0$
Thus,the required equation is $x^2 + y^2 + z^2 - 2a(x + y + z) + 2a^2 = 0$.

(C) The diameter $d$ of the sphere is the distance between the two given end points $(3, 4, -1)$ and $(-1, 2, 3)$.
Using the distance formula,$d = \sqrt{(-1 - 3)^2 + (2 - 4)^2 + (3 - (-1))^2}$.
$d = \sqrt{(-4)^2 + (-2)^2 + (4)^2}$.
$d = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
The radius $r$ of the sphere is half of the diameter.
$r = \frac{d}{2} = \frac{6}{2} = 3$.
Thus,the radius of the sphere is $3$.

(C) The required point is the center of the sphere passing through the given four points.
Let the equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \dots (i)$.
Since the sphere passes through $(0, 0, 0)$,we get $d = 0$.
Since it passes through $(a, 0, 0)$,we have $a^2 + 2ua = 0 \implies u = -a/2$.
Since it passes through $(0, b, 0)$,we have $b^2 + 2vb = 0 \implies v = -b/2$.
Since it passes through $(0, 0, c)$,we have $c^2 + 2wc = 0 \implies w = -c/2$.
The center of the sphere is $(-u, -v, -w) = (a/2, b/2, c/2)$.
This point is equidistant from all four given points.

(D) sphere of radius $r$ that touches all three coordinate axes must have its center at a point $(\pm r, \pm r, \pm r)$.
Since there are $8$ octants in the three-dimensional Cartesian coordinate system,each octant allows for one such sphere.
Therefore,there are $8$ different spheres of radius $r$ that can be drawn to touch all three coordinate axes.

(A) Let the coordinates of $A, B,$ and $C$ be $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.
The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane passes through the fixed point $(p, q, r)$,we have $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$.
The equation of the sphere passing through the origin $O(0, 0, 0)$ and points $A, B, C$ is given by $x^2 + y^2 + z^2 - ax - by - cz = 0$.
The centre of this sphere is $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$.
Let the centre be $(x, y, z)$. Then $x = \frac{a}{2}, y = \frac{b}{2}, z = \frac{c}{2}$,which implies $a = 2x, b = 2y, c = 2z$.
Substituting these values into the plane equation $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$,we get $\frac{p}{2x} + \frac{q}{2y} + \frac{r}{2z} = 1$.
Multiplying by $2$,the locus of the centre is $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 2$.

(A) Let the ratio in which the sphere divides the line segment $AB$ be $\lambda : 1$.
Using the section formula,the coordinates of the point $P$ on the line segment $AB$ are given by:
$P = \left( \frac{27\lambda + 12}{\lambda + 1}, \frac{-9\lambda - 4}{\lambda + 1}, \frac{18\lambda + 8}{\lambda + 1} \right)$.
Since the point $P$ lies on the sphere ${x^2} + {y^2} + {z^2} = 504$,we substitute these coordinates into the sphere equation:
$\left( \frac{27\lambda + 12}{\lambda + 1} \right)^2 + \left( \frac{-9\lambda - 4}{\lambda + 1} \right)^2 + \left( \frac{18\lambda + 8}{\lambda + 1} \right)^2 = 504$.
Factoring out the common terms:
$\frac{1}{(\lambda + 1)^2} [ (3(9\lambda + 4))^2 + (-(9\lambda + 4))^2 + (2(9\lambda + 4))^2 ] = 504$.
$\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} [ 9 + 1 + 4 ] = 504$.
$\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} \times 14 = 504$.
$\frac{(9\lambda + 4)^2}{(\lambda + 1)^2} = 36$.
Taking the square root on both sides:
$\frac{9\lambda + 4}{\lambda + 1} = \pm 6$.
Case $1$: $9\lambda + 4 = 6\lambda + 6 \implies 3\lambda = 2 \implies \lambda = 2/3$ (Internal division).
Case $2$: $9\lambda + 4 = -6\lambda - 6 \implies 15\lambda = -10 \implies \lambda = -2/3$ (External division).
Since the option $2:3$ externally is provided,the correct answer is $A$.

(D) Let the two spheres have centers $O$ and $O'$ and radii $r_1$ and $r_2$ respectively.
Since the spheres cut orthogonally,the angle between the radii at the point of intersection $P$ is $90^\circ$.
Let $C$ be the center of the common circle and $r$ be its radius.
In $\Delta OPC$,the angle $\angle OPC = \theta$. Thus,$\cos \theta = \frac{r}{r_1}$.
In $\Delta O'PC$,the angle $\angle O'PC = 90^\circ - \theta$. Thus,$\sin \theta = \cos(90^\circ - \theta) = \frac{r}{r_2}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\left(\frac{r}{r_1}\right)^2 + \left(\frac{r}{r_2}\right)^2 = 1$
$r^2 \left(\frac{1}{r_1^2} + \frac{1}{r_2^2}\right) = 1$
$r^2 \left(\frac{r_1^2 + r_2^2}{r_1^2 r_2^2}\right) = 1$
$r^2 = \frac{r_1^2 r_2^2}{r_1^2 + r_2^2}$
$r = \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}}$

(A) The equation of the first sphere is ${S_1} \equiv {x^2} + {y^2} + {z^2} + 6x - 8y - 2z - 13 = 0$. Its center ${C_1}$ is $(-3, 4, 1)$.
The equation of the second sphere is ${S_2} \equiv {x^2} + {y^2} + {z^2} - 10x + 4y - 2z - 8 = 0$. Its center ${C_2}$ is $(5, -2, 1)$.
The midpoint $P$ of the line segment joining ${C_1}$ and ${C_2}$ is given by:
$P = \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = (1, 1, 1)$.
Since the plane $2ax - 3ay + 4az + 6 = 0$ passes through the point $P(1, 1, 1)$,we substitute the coordinates of $P$ into the plane equation:
$2a(1) - 3a(1) + 4a(1) + 6 = 0$
$2a - 3a + 4a + 6 = 0$
$3a + 6 = 0$
$3a = -6$
$a = -2$.

(D) The given equation of the sphere is $x^2 + y^2 + z^2 - x + z - 2 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get $u = -1/2$,$v = 0$,$w = 1/2$,and $d = -2$.
The center of the sphere is $(-u, -v, -w) = (1/2, 0, -1/2)$.
The radius of the sphere $R$ is given by $\sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1/2)^2 + 0^2 + (1/2)^2 - (-2)} = \sqrt{1/4 + 0 + 1/4 + 2} = \sqrt{1/2 + 2} = \sqrt{5/2}$.
The perpendicular distance $P$ from the center $(1/2, 0, -1/2)$ to the plane $x + 2y - z - 4 = 0$ is:
$P = \frac{|(1/2) + 2(0) - (-1/2) - 4|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|1/2 + 1/2 - 4|}{\sqrt{6}} = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}}$.
The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - P^2}$.
$r = \sqrt{\frac{5}{2} - \frac{3}{2}} = \sqrt{\frac{2}{2}} = \sqrt{1} = 1$.

(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 2y - 4z = 11$.
Rewriting it as ${x^2} + {(y - 1)^2} + {(z - 2)^2} = 11 + 1 + 4 = 16$.
Thus,the centre of the sphere is $C = (0, 1, 2)$ and the radius $R = 4$.
The distance $d$ from the centre of the sphere to the plane $x + 2y + 2z - 15 = 0$ is given by $d = \frac{|0 + 2(1) + 2(2) - 15|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|2 + 4 - 15|}{\sqrt{9}} = \frac{|-9|}{3} = 3$.
The radius $r$ of the circle is given by $r = \sqrt{R^2 - d^2}$.
Substituting the values,$r = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7}$.

(D) Given the equation of the sphere: $2x^2 + 2y^2 + 2z^2 - 6x + 2y - 4z = 1$.
Dividing by $2$,we get: $x^2 + y^2 + z^2 - 3x + y - 2z = \frac{1}{2}$.
The center of the sphere is $C = (\frac{3}{2}, -\frac{1}{2}, 1)$.
The radius $r$ is given by $\sqrt{g^2 + f^2 + h^2 - d} = \sqrt{(\frac{3}{2})^2 + (-\frac{1}{2})^2 + (1)^2 - (-\frac{1}{2})} = \sqrt{\frac{9}{4} + \frac{1}{4} + 1 + \frac{1}{2}} = \sqrt{\frac{16}{4}} = 2$.
The required sphere is concentric,so its center is also $(\frac{3}{2}, -\frac{1}{2}, 1)$.
The radius of the required sphere is $R = 2 \times r = 2 \times 2 = 4$.
The equation of the required sphere is $(x - \frac{3}{2})^2 + (y + \frac{1}{2})^2 + (z - 1)^2 = 4^2$.
Expanding this: $x^2 - 3x + \frac{9}{4} + y^2 + y + \frac{1}{4} + z^2 - 2z + 1 = 16$.
$x^2 + y^2 + z^2 - 3x + y - 2z + \frac{14}{4} = 16$.
$x^2 + y^2 + z^2 - 3x + y - 2z + 3.5 - 16 = 0$.
$x^2 + y^2 + z^2 - 3x + y - 2z - 12.5 = 0$.
Multiplying by $2$: $2x^2 + 2y^2 + 2z^2 - 6x + 2y - 4z - 25 = 0$.

(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0$.
Comparing this with the general form ${x^2} + {y^2} + {z^2} + 2ux + 2vy + 2wz + d = 0$,we get the center of the sphere as $(-u, -v, -w) = (3, 6, 1)$.
Let the coordinates of the other end of the diameter be $(x, y, z)$.
Since the center of the sphere is the midpoint of the diameter,we have:
$\frac{2 + x}{2} = 3 \implies 2 + x = 6 \implies x = 4$
$\frac{3 + y}{2} = 6 \implies 3 + y = 12 \implies y = 9$
$\frac{5 + z}{2} = 1 \implies 5 + z = 2 \implies z = -3$
Thus,the coordinates of the other end are $(4, 9, -3)$.

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 4x - 2y - 6z - 155 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2gx + 2fy + 2hz + c = 0$,we get the center $C = (-g, -f, -h) = (-2, 1, 3)$.
The radius $r$ is given by $\sqrt{g^2 + f^2 + h^2 - c} = \sqrt{(-2)^2 + 1^2 + (-3)^2 - (-155)} = \sqrt{4 + 1 + 9 + 155} = \sqrt{169} = 13$.
The perpendicular distance $d$ from the center $C(-2, 1, 3)$ to the plane $12x + 4y + 3z - 327 = 0$ is calculated as:
$d = \frac{|12(-2) + 4(1) + 3(3) - 327|}{\sqrt{12^2 + 4^2 + 3^2}} = \frac{|-24 + 4 + 9 - 327|}{\sqrt{144 + 16 + 9}} = \frac{|-338|}{\sqrt{169}} = \frac{338}{13} = 26$.
The shortest distance from the plane to the sphere is $d - r = 26 - 13 = 13$.

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get the center $C = (-u, -v, -w) = (-1, 1, 2)$ and the radius $R = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1 + 1 + 4 + 19} = \sqrt{25} = 5$.
The perpendicular distance $p$ from the center $C(-1, 1, 2)$ to the plane $x + 2y + 2z + 7 = 0$ is given by $p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1 + 4 + 4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - p^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.

(B) The given equation of the sphere is $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$.
Rewriting it in standard form: $x^{2} + (y-1)^{2} - 1 + (z-2)^{2} - 4 = 11$,which simplifies to $x^{2} + (y-1)^{2} + (z-2)^{2} = 16$.
The center of the sphere is $C(0, 1, 2)$ and its radius $R = \sqrt{16} = 4$.
The distance $d$ from the center $C(0, 1, 2)$ to the plane $x + 2y + 2z - 15 = 0$ is given by $d = \frac{|0 + 2(1) + 2(2) - 15|}{\sqrt{1^{2} + 2^{2} + 2^{2}}} = \frac{|2 + 4 - 15|}{\sqrt{1 + 4 + 4}} = \frac{|-9|}{3} = 3$.
The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^{2} - d^{2}}$.
Substituting the values,$r = \sqrt{4^{2} - 3^{2}} = \sqrt{16 - 9} = \sqrt{7}$.

(A) Let the equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$.
Since it passes through the origin $(0, 0, 0)$,we have $d = 0$.
Since it passes through $(0, 2, 0)$,we have $0^2 + 2^2 + 0^2 + 2u(0) + 2v(2) + 2w(0) + 0 = 0$,which gives $4 + 4v = 0$,so $v = -1$.
Since it passes through $(1, 0, 0)$,we have $1^2 + 0^2 + 0^2 + 2u(1) + 2v(0) + 2w(0) + 0 = 0$,which gives $1 + 2u = 0$,so $u = -1/2$.
Since it passes through $(0, 0, 4)$,we have $0^2 + 0^2 + 4^2 + 2u(0) + 2v(0) + 2w(4) + 0 = 0$,which gives $16 + 8w = 0$,so $w = -2$.
The center of the sphere is $(-u, -v, -w) = \left( -(-1/2), -(-1), -(-2) \right) = \left( \frac{1}{2}, 1, 2 \right)$.

(A) The equation of the first sphere is $x^2 + y^2 + z^2 + 6x - 8y - 2z - 13 = 0$. Its center $C_1$ is $(-3, 4, 1)$.
The equation of the second sphere is $x^2 + y^2 + z^2 - 10x + 4y - 2z - 8 = 0$. Its center $C_2$ is $(5, -2, 1)$.
The midpoint $P$ of the line segment joining $C_1$ and $C_2$ is given by $P = \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = (1, 1, 1)$.
Since the plane $2ax - 3ay + 4az + 6 = 0$ passes through $P(1, 1, 1)$,we substitute the coordinates of $P$ into the plane equation:
$2a(1) - 3a(1) + 4a(1) + 6 = 0$
$2a - 3a + 4a + 6 = 0$
$3a + 6 = 0$
$3a = -6$
$a = -2$.

(D) sphere touching all three coordinate planes ($xy$,$yz$,and $zx$) must have its center at a point $(\pm r, \pm r, \pm r)$.
Since there are $3$ coordinate planes and the sphere must be tangent to all of them,the distance from the center to each plane must be equal to the radius $r$.
There are $2^3 = 8$ possible combinations of signs for the coordinates of the center $(\pm r, \pm r, \pm r)$.
Thus,there are $8$ distinct spheres of radius $r$ that touch all three coordinate planes,one in each octant of the $3D$ coordinate system.

(A) Let the moving point be $P(x, y, z)$.
Given that the distance of point $P$ from the fixed point $C(1, -2, 2)$ is $1$.
Using the distance formula,we have $\sqrt{(x - 1)^2 + (y + 2)^2 + (z - 2)^2} = 1$.
Squaring both sides,we get $(x - 1)^2 + (y + 2)^2 + (z - 2)^2 = 1^2$.
Expanding the squares: $(x^2 - 2x + 1) + (y^2 + 4y + 4) + (z^2 - 4z + 4) = 1$.
Simplifying the equation: $x^2 + y^2 + z^2 - 2x + 4y - 4z + 9 = 1$.
Subtracting $1$ from both sides: $x^2 + y^2 + z^2 - 2x + 4y - 4z + 8 = 0$.

(D) The equation of the given sphere is $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get $2u = -6, 2v = -12, 2w = -2$.
Thus,$u = -3, v = -6, w = -1$.
The center of the sphere is $(-u, -v, -w) = (3, 6, 1)$.
Let the given end of the diameter be $A = (2, 3, 5)$ and the other end be $B = (\alpha, \beta, \gamma)$.
Since the center of the sphere is the midpoint of the diameter,we have:
$\frac{\alpha + 2}{2} = 3 \Rightarrow \alpha + 2 = 6 \Rightarrow \alpha = 4$
$\frac{\beta + 3}{2} = 6 \Rightarrow \beta + 3 = 12 \Rightarrow \beta = 9$
$\frac{\gamma + 5}{2} = 1 \Rightarrow \gamma + 5 = 2 \Rightarrow \gamma = -3$
Therefore,the coordinates of the other end are $(4, 9, -3)$.

(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.
Given the spheres:
${S_1}: {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$
${S_2}: {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$
Subtracting the two equations:
$({x^2} + {y^2} + {z^2} + 7x - 2y - z - 13) - ({x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8) = 0$
$(7x - (-3x)) + (-2y - 3y) + (-z - 4z) + (-13 - (-8)) = 0$
$10x - 5y - 5z - 5 = 0$
Dividing the entire equation by $5$:
$2x - y - z = 1$
Thus,the intersection of the two spheres lies on the plane $2x - y - z = 1$.

(D) The given equation of the sphere is $x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0$.
Completing the squares,we get $(x - 1)^2 + (y + 2)^2 + (z - 3)^2 = -10 + 1 + 4 + 9 = 4$.
Thus,the center of the sphere is $C(1, -2, 3)$ and the radius $r = \sqrt{4} = 2$.
Now,find the perpendicular distance $d$ from the center $C(1, -2, 3)$ to the plane $2x + y - 2z - 6 = 0$ using the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
$d = \frac{|2(1) + 1(-2) - 2(3) - 6|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|2 - 2 - 6 - 6|}{\sqrt{4 + 1 + 4}} = \frac{|-12|}{\sqrt{9}} = \frac{12}{3} = 4$.
The least distance between the sphere and the plane is given by $d - r = 4 - 2 = 2$ units.

Let the coordinates of point $P$ be $(x, y, z)$.
Given $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.
Using the distance formula,$PA^{2} = (x-3)^{2} + (y-4)^{2} + (z-5)^{2} = x^{2} - 6x + 9 + y^{2} - 8y + 16 + z^{2} - 10z + 25 = x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50$.
Similarly,$PB^{2} = (x+1)^{2} + (y-3)^{2} + (z+7)^{2} = x^{2} + 2x + 1 + y^{2} - 6y + 9 + z^{2} + 14z + 49 = x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59$.
Given the condition $PA^{2} + PB^{2} = k^{2}$,we have:
$(x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50) + (x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59) = k^{2}$.
Combining like terms:
$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z + 109 = k^{2}$.
Dividing by $2$:
$x^{2} + y^{2} + z^{2} - 2x - 7y + 2z + \frac{109}{2} = \frac{k^{2}}{2}$.
Thus,the equation is $x^{2} + y^{2} + z^{2} - 2x - 7y + 2z = \frac{k^{2} - 109}{2}$.

(A) Let the radius of the sphere be $r$. Since the sphere touches the two perpendicular walls and the floor,we can set up a coordinate system where the walls and floor are the coordinate planes $x=0, y=0, z=0$. The center of the sphere is at $(r, r, r)$.
The equation of the sphere is $(x-r)^2 + (y-r)^2 + (z-r)^2 = r^2$.
$A$ point on the sphere is given at distances $9, 16, 25$ from the walls and floor,so the coordinates of this point are $(9, 16, 25)$.
Substituting this point into the sphere's equation:
$(9-r)^2 + (16-r)^2 + (25-r)^2 = r^2$
Expanding the terms:
$(81 - 18r + r^2) + (256 - 32r + r^2) + (625 - 50r + r^2) = r^2$
$3r^2 - 100r + 962 = r^2$
$2r^2 - 100r + 962 = 0$
Dividing by $2$:
$r^2 - 50r + 481 = 0$
Factoring the quadratic equation:
$(r-13)(r-37) = 0$
Thus,$r = 13$ or $r = 37$.
The possible radius is $13$.

(B) The given equation of the sphere is $(x-2)^2+(y-3)^2+(z-6)^2=1$.
This represents a sphere with center $C = (2, 3, 6)$ and radius $r = 1$.
The distance $d$ from the origin $O(0, 0, 0)$ to the center $C(2, 3, 6)$ is calculated as:
$d = \sqrt{(2-0)^2 + (3-0)^2 + (6-0)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
The shortest distance from the origin to any point on the sphere is given by the distance from the origin to the center minus the radius of the sphere.
Shortest distance $= d - r = 7 - 1 = 6$.

(B) Let the initial radius of the sphere be $r$ and the initial volume be $V = \frac{4}{3} \pi r^3$.
After the increase,the new volume $V'$ is $V + 0.728V = 1.728V$.
Since $V' = \frac{4}{3} \pi (r')^3$,we have $\frac{4}{3} \pi (r')^3 = 1.728 \times \frac{4}{3} \pi r^3$.
Thus,$(r')^3 = 1.728 r^3$,which implies $r' = \sqrt[3]{1.728} r = 1.2r$.
The initial surface area is $S = 4 \pi r^2$ and the new surface area is $S' = 4 \pi (r')^2$.
$S' = 4 \pi (1.2r)^2 = 4 \pi (1.44 r^2) = 1.44 S$.
The percentage increase in surface area is $\frac{S' - S}{S} \times 100 = (1.44 - 1) \times 100 = 44 \%$.

(B) The angle subtended by the diameter at any point on the circle is a right angle. Therefore,$\angle APB = 90^{\circ}$,which implies $AP \perp PB$.
The direction ratios of $AP$ are $(5-3, 6-(-2), -1-2) = (2, 8, -3)$.
The direction ratios of $PB$ are $(2-5, \lambda+1-6, 5-(-1)) = (-3, \lambda-5, 6)$.
Since $AP \perp PB$,the dot product of their direction ratios must be zero:
$(2)(-3) + (8)(\lambda-5) + (-3)(6) = 0$
$-6 + 8\lambda - 40 - 18 = 0$
$8\lambda - 64 = 0$
$8\lambda = 64$
$\lambda = 8$.

(C) Let the point be $P(x, y, z)$. The points are $A(-3, 1, 2)$ and $B(1, -2, 4)$.
Since the line segment $AB$ subtends a right angle at $P$,the vectors $\vec{PA}$ and $\vec{PB}$ are perpendicular.
$\vec{PA} = (-3-x, 1-y, 2-z)$
$\vec{PB} = (1-x, -2-y, 4-z)$
Since $\vec{PA} \cdot \vec{PB} = 0$,we have:
$(-3-x)(1-x) + (1-y)(-2-y) + (2-z)(4-z) = 0$
$(x+3)(x-1) + (y-1)(y+2) + (z-2)(z-4) = 0$
$(x^2 + 2x - 3) + (y^2 + y - 2) + (z^2 - 6z + 8) = 0$
$x^2 + y^2 + z^2 + 2x + y - 6z + (-3 - 2 + 8) = 0$
$x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$
Thus,the locus is $x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$.

(A) Let the points be $A(1,0,0), B(0,1,0)$ and $C(1,1,1)$.
Calculate the distances between the points:
$AB = \sqrt{(0-1)^2 + (1-0)^2 + 0^2} = \sqrt{2}$
$BC = \sqrt{(1-0)^2 + (1-1)^2 + (1-0)^2} = \sqrt{2}$
$CA = \sqrt{(1-1)^2 + (1-0)^2 + (1-0)^2} = \sqrt{2}$
Since $AB = BC = CA = \sqrt{2}$,the points form an equilateral triangle.
The sphere with the smallest radius passing through these points has its center at the centroid of the triangle $ABC$.
Center $C' = \left(\frac{1+0+1}{3}, \frac{0+1+1}{3}, \frac{0+0+1}{3}\right) = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)$.
The radius $R$ is the distance from $C'$ to any point,say $A(1,0,0)$:
$R^2 = \left(\frac{2}{3}-1\right)^2 + \left(\frac{2}{3}-0\right)^2 + \left(\frac{1}{3}-0\right)^2 = \frac{1}{9} + \frac{4}{9} + \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$.
The equation of the sphere is $(x-\frac{2}{3})^2 + (y-\frac{2}{3})^2 + (z-\frac{1}{3})^2 = \frac{2}{3}$.
Expanding this: $x^2 - \frac{4x}{3} + \frac{4}{9} + y^2 - \frac{4y}{3} + \frac{4}{9} + z^2 - \frac{2z}{3} + \frac{1}{9} = \frac{6}{9}$.
$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + \frac{9}{9} = \frac{6}{9}$.
$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + 1 = \frac{2}{3}$.
Multiplying by $3$: $3(x^2+y^2+z^2) - 4x - 4y - 2z + 3 = 2$.
$3(x^2+y^2+z^2) - 4x - 4y - 2z + 1 = 0$.

(B) The equation of the sphere is $x^2+y^2+z^2+2x-2y-4z-19=0$.
Comparing this with the general equation $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get the centre $C = (-u, -v, -w) = (-1, 1, 2)$.
The radius of the sphere $R$ is given by $R = \sqrt{u^2+v^2+w^2-d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$.
The perpendicular distance $p$ from the centre $C(-1, 1, 2)$ to the plane $x+2y+2z+7=0$ is:
$p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1+4+4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
Let $r$ be the radius of the circle. In the right-angled triangle formed by the centre of the sphere,the centre of the circle,and a point on the circle,we have $R^2 = p^2 + r^2$.
$r^2 = R^2 - p^2 = 5^2 - 4^2 = 25 - 16 = 9$.
Therefore,$r = \sqrt{9} = 3$.

(B) The equation of the sphere is $x^2+y^2+z^2-6x-12y-2z+20=0$.
Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-6, 2v=-12, 2w=-2$.
Thus,$u=-3, v=-6, w=-1$.
The centre of the sphere is $(-u,-v,-w) = (3,6,1)$.
Let the given end of the diameter be $A = (2,3,-3)$ and the other end be $B = (\alpha, \beta, \gamma)$.
Since the centre $O(3,6,1)$ is the midpoint of the diameter $AB$,we have:
$O = \left( \frac{\alpha+2}{2}, \frac{\beta+3}{2}, \frac{\gamma-3}{2} \right) = (3,6,1)$.
Equating the coordinates:
$\frac{\alpha+2}{2} = 3 \Rightarrow \alpha+2 = 6 \Rightarrow \alpha = 4$.
$\frac{\beta+3}{2} = 6 \Rightarrow \beta+3 = 12 \Rightarrow \beta = 9$.
$\frac{\gamma-3}{2} = 1 \Rightarrow \gamma-3 = 2 \Rightarrow \gamma = 5$.
Therefore,the other end of the diameter is $(4,9,5)$.

(A) The given equation of the sphere is $x^2+y^2+z^2-12x-4y-3z=0$.
Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-12$,$2v=-4$,and $2w=-3$.
Thus,$u=-6$,$v=-2$,and $w=-\frac{3}{2}$.
The centre of the sphere is $(-u, -v, -w) = (6, 2, \frac{3}{2})$.
The radius $r$ of the sphere is given by the formula $r = \sqrt{u^2+v^2+w^2-d}$.
Substituting the values,we get $r = \sqrt{(-6)^2+(-2)^2+(-\frac{3}{2})^2-0}$.
$r = \sqrt{36+4+\frac{9}{4}} = \sqrt{40+\frac{9}{4}} = \sqrt{\frac{160+9}{4}} = \sqrt{\frac{169}{4}}$.
Therefore,$r = \frac{13}{2}$.

(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $a, b, c$ are the direction ratios of the normal to the plane.
Let $Q(x_1, y_1, z_1)$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane.
Since $OQ$ is perpendicular to the plane,the direction ratios of $OQ$ are $(x_1, y_1, z_1)$,which must be proportional to $(a, b, c)$.
Thus,$a = kx_1, b = ky_1, c = kz_1$ for some constant $k$.
Substituting these into the plane equation: $x_1(x_1 - 1) + y_1(y_1 - 2) + z_1(z_1 - 3) = 0$.
This simplifies to $x_1^2 + y_1^2 + z_1^2 - x_1 - 2y_1 - 3z_1 = 0$.
This is the equation of a sphere with diameter $OP$,where $O$ is the origin and $P$ is the fixed point $(1, 2, 3)$.

(B) The equation of the sphere is $x^2+y^2+z^2+2x-2y-4z-19=0$. Comparing this with the general equation $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $u=1, v=-1, w=-2, d=-19$.
The centre of the sphere is $(-u, -v, -w) = (-1, 1, 2)$.
The radius of the sphere $R$ is given by $\sqrt{u^2+v^2+w^2-d} = \sqrt{1^2+(-1)^2+(-2)^2-(-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$.
Now,the perpendicular distance $p$ from the centre $(-1, 1, 2)$ to the plane $x+2y+2z+7=0$ is:
$p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2+2^2+2^2}} = \frac{|-1+2+4+7|}{\sqrt{1+4+4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
Let $r$ be the radius of the circle. In the right-angled triangle formed by the centre of the sphere,the centre of the circle,and a point on the circumference of the circle,we have $R^2 = p^2 + r^2$.
$r^2 = R^2 - p^2 = 5^2 - 4^2 = 25 - 16 = 9$.
Therefore,$r = \sqrt{9} = 3$.