AP EAMCET 2019 Mathematics Question Paper with Answer and Solution

(D) Given the equation for $x > 2$:
$\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$
Squaring both sides,we get:
$(x+2) + (x-2) - 2\sqrt{(x+2)(x-2)} = 4x - 2$
$2x - 2\sqrt{x^2-4} = 4x - 2$
$-2\sqrt{x^2-4} = 2x - 2$
$-\sqrt{x^2-4} = x - 1$
Squaring again:
$x^2 - 4 = (x-1)^2$
$x^2 - 4 = x^2 - 2x + 1$
$-4 = -2x + 1$
$2x = 5$
$x = 2.5$
Now,check the value $x = 2.5$ in the original equation:
$LHS$: $\sqrt{2.5+2} - \sqrt{2.5-2} = \sqrt{4.5} - \sqrt{0.5} = \sqrt{9 \times 0.5} - \sqrt{0.5} = 3\sqrt{0.5} - \sqrt{0.5} = 2\sqrt{0.5} = \sqrt{4 \times 0.5} = \sqrt{2}$
$RHS$: $\sqrt{4(2.5) - 2} = \sqrt{10 - 2} = \sqrt{8} = 2\sqrt{2}$
Since $LHS$ $\neq$ $RHS$,there is no solution. Thus,option $D$ is correct.

(C) Given equation: $\sin x \sqrt{4 \cos^2 x} = \frac{2+x-[x]}{1-x+[x]}$.
Since $x-[x] = \{x\}$,the equation becomes $\sin x \cdot 2|\cos x| = \frac{2+\{x\}}{1-\{x\}}$.
For $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,$\cos x > 0$,so $|\cos x| = \cos x$.
The equation simplifies to $\sin 2x = \frac{2+\{x\}}{1-\{x\}}$.
In the interval $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,the maximum value of $\sin 2x$ is $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866$.
The expression $f(\{x\}) = \frac{2+\{x\}}{1-\{x\}}$ for $\{x\} \in [0, 1)$ has a minimum value at $\{x\} = 0$,which is $f(0) = \frac{2+0}{1-0} = 2$.
Since the maximum value of the $LHS$ $(0.866)$ is less than the minimum value of the $RHS$ $(2)$,there are no solutions for $x$ in the given interval.
Thus,$k = 0$.
Therefore,$k^{\tan^2 x} = 0^{\tan^2 x} = 0$ for all $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$.

(B) Given $f(\theta) = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} + \sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}$.
Squaring both sides,we get $[f(\theta)]^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta) + 2 \sqrt{(a^2 \cos^2 \theta + b^2 \sin^2 \theta)(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}$.
Simplifying the terms,$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^4 \sin^2 \theta \cos^2 \theta + a^2 b^2 \cos^4 \theta + a^2 b^2 \sin^4 \theta + b^4 \sin^2 \theta \cos^2 \theta}$.
$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (\sin^4 \theta + \cos^4 \theta) + (a^4 + b^4) \sin^2 \theta \cos^2 \theta}$.
Using $\sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$,we get $[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (1 - 2 \sin^2 \theta \cos^2 \theta) + (a^4 + b^4) \sin^2 \theta \cos^2 \theta}$.
$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^4 + b^4 - 2 a^2 b^2) \sin^2 \theta \cos^2 \theta} = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^2 - b^2)^2 \sin^2 \theta \cos^2 \theta}$.
Let $X = \sin^2 \theta \cos^2 \theta = \frac{\sin^2(2\theta)}{4}$. The range of $X$ is $[0, 1/4]$.
For maximum value $M$,$X = 1/4$: $M = a^2 + b^2 + 2 \sqrt{a^2 b^2 + \frac{(a^2 - b^2)^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{4a^2 b^2 + a^4 + b^4 - 2a^2 b^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{(a^2 + b^2)^2}{4}} = a^2 + b^2 + (a^2 + b^2) = 2(a^2 + b^2)$.
For minimum value $m$,$X = 0$: $m = a^2 + b^2 + 2 \sqrt{a^2 b^2} = a^2 + b^2 + 2ab = (a+b)^2$.
Thus,$M - m = 2a^2 + 2b^2 - (a^2 + b^2 + 2ab) = a^2 + b^2 - 2ab = (a-b)^2$.

(B) Given,$f(x) = \tan \left(x + \frac{2 \pi}{3} \right) - \tan \left(x + \frac{\pi}{6} \right) + \cos \left(x + \frac{\pi}{6} \right)$.
Using the identity $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$,we have:
$\tan \left(x + \frac{2 \pi}{3} \right) - \tan \left(x + \frac{\pi}{6} \right) = \frac{\sin \left( \frac{2 \pi}{3} - \frac{\pi}{6} \right)}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)} = \frac{\sin \frac{\pi}{2}}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)} = \frac{1}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)}$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,the expression becomes:
$f(x) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} \right) + \cos \frac{\pi}{2}} + \cos \left(x + \frac{\pi}{6} \right) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} \right)} + \cos \left(x + \frac{\pi}{6} \right)$.
In the interval $\left[ -\frac{5 \pi}{12}, -\frac{\pi}{3} \right]$,the function is monotonically increasing.
Thus,the maximum value occurs at the right endpoint $x = -\frac{\pi}{3} = -60^{\circ}$.
$f(-60^{\circ}) = \tan \left( -60^{\circ} + 120^{\circ} \right) - \tan \left( -60^{\circ} + 30^{\circ} \right) + \cos \left( -60^{\circ} + 30^{\circ} \right) = \tan 60^{\circ} - \tan(-30^{\circ}) + \cos(-30^{\circ}) = \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \sqrt{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2} = \frac{6 \sqrt{3} + 2 \sqrt{3} + 3 \sqrt{3}}{6} = \frac{11 \sqrt{3}}{6}$.

(D) The maximum value of the quadratic expression $1-2x-5x^2$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(-5)} = -\frac{1}{5}$.
Substituting $x = -\frac{1}{5}$,we get $a = 1 - 2(-\frac{1}{5}) - 5(-\frac{1}{5})^2 = 1 + \frac{2}{5} - \frac{1}{5} = \frac{6}{5}$.
The minimum value of the quadratic expression $x^2-2x+5$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$.
Substituting $x = 1$,we get $b = (1)^2 - 2(1) + 5 = 4$.
Now,substitute $a = \frac{6}{5}$ and $b = 4$ into the expression $5ax^2+bx+7$:
$5(\frac{6}{5})x^2 + 4x + 7 = 6x^2 + 4x + 7$.
For the quadratic $6x^2 + 4x + 7 > 0$,we check the discriminant $D = b^2 - 4ac = (4)^2 - 4(6)(7) = 16 - 168 = -152$.
Since $D < 0$ and the leading coefficient $6 > 0$,the expression $6x^2 + 4x + 7$ is always positive for all real values of $x$.
Thus,the set of all values of $x$ is $(-\infty, \infty)$.

(D) Given the expression:
$\frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$
Since the degree of the numerator is $4$ and the degree of the denominator is $3$,$P(x)$ must be a linear polynomial of the form $P(x) = x+k$.
Comparing the coefficients of $x^4$ on both sides,we see that $P(x) = x+k$.
Multiplying by $(x-a)(x-b)(x-c)$,we get:
$x^4 = (x-a)(x-b)(x-c)P(x) + A(x-b)(x-c) + B(x-a)(x-c) + C(x-a)(x-b)$
Setting $x=a$,we get $A = \frac{a^4}{(a-b)(a-c)}$.
Thus,$A(a-b)(a-c) = a^4$.
To find $P(0)$,we substitute $x=0$ in the original equation:
$\frac{0}{(-a)(-b)(-c)} = P(0) + \frac{A}{-a} + \frac{B}{-b} + \frac{C}{-c}$
$0 = P(0) - (\frac{A}{a} + \frac{B}{b} + \frac{C}{c})$
$P(0) = \frac{A}{a} + \frac{B}{b} + \frac{C}{c}$.
Using the values $A = \frac{a^4}{(a-b)(a-c)}$,$B = \frac{b^4}{(b-a)(b-c)}$,and $C = \frac{c^4}{(c-a)(c-b)}$,we find $P(0) = a+b+c$.
Therefore,$P(0) + A(a-b)(a-c) = (a+b+c) + a^4$.
Hence,the correct option is $D$.

(C) Given,$\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$
Multiplying both sides by $(x+3)^2(x-2)$,we get:
$8 = (Ax+B)(x-2) + C(x+3)^2$
Setting $x=2$:
$8 = 0 + C(2+3)^2 \Rightarrow 8 = 25C \Rightarrow C = \frac{8}{25}$
Setting $x=-3$:
$8 = (A(-3)+B)(-3-2) + 0 \Rightarrow 8 = (-3A+B)(-5) \Rightarrow 8 = 15A - 5B$
Comparing coefficients of $x^2$:
$0 = A + C \Rightarrow A = -C = -\frac{8}{25}$
Substituting $A$ into $8 = 15A - 5B$:
$8 = 15(-\frac{8}{25}) - 5B \Rightarrow 8 = -\frac{24}{5} - 5B \Rightarrow 5B = -\frac{24}{5} - 8 = -\frac{64}{5} \Rightarrow B = -\frac{64}{25}$
Now,calculate $25(B+8C-A)$:
$25(-\frac{64}{25} + 8(\frac{8}{25}) - (-\frac{8}{25})) = 25(-\frac{64}{25} + \frac{64}{25} + \frac{8}{25}) = 25(\frac{8}{25}) = 8$

(C) Let $X = \frac{\sqrt{3}x - y}{2}$ and $Y = \frac{x + \sqrt{3}y}{2}$.
This transformation represents a rotation of the coordinate axes by an angle $\theta = 30^\circ$ (or $\pi/6$ radians).
Since rotation is an isometry (it preserves distances and areas),the area of the region bounded by the curves in the new coordinate system $(X, Y)$ is identical to the area of the region bounded by the curves $Y = X^2$ and $X = Y^2$ in the original coordinate system $(x, y)$.
The given equations $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ transform into $Y = X^2$ and $X = Y^2$ respectively.
Therefore,the area of the region bounded by these curves is equal to the area of the region bounded by $y=x^2$ and $x=y^2$,which is given as $k$.

(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.
Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.
Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.
Equating the midpoints:
$\frac{3+x}{2} = 3 \implies 3+x = 6 \implies x = 3$.
$\frac{6+y}{2} = \frac{9}{2} \implies 6+y = 9 \implies y = 3$.
$\frac{-1+z}{2} = 0 \implies -1+z = 0 \implies z = 1$.
Thus,the fourth vertex $D$ is $(3, 3, 1)$.
Therefore,option $(D)$ is correct.

(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(1-2)^2 + (0-(-1))^2 + (-1-4)^2} = \sqrt{(-1)^2 + 1^2 + (-5)^2} = \sqrt{1+1+25} = \sqrt{27}$.
$BC = \sqrt{(1-1)^2 + (2-0)^2 + (3-(-1))^2} = \sqrt{0^2 + 2^2 + 4^2} = \sqrt{0+4+16} = \sqrt{20}$.
$CD = \sqrt{(2-1)^2 + (1-2)^2 + (8-3)^2} = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1+1+25} = \sqrt{27}$.
$DA = \sqrt{(2-2)^2 + (-1-1)^2 + (4-8)^2} = \sqrt{0^2 + (-2)^2 + (-4)^2} = \sqrt{0+4+16} = \sqrt{20}$.
Since $AB = CD = \sqrt{27}$ and $BC = DA = \sqrt{20}$,the opposite sides are equal.
Now,we check the diagonals $AC$ and $BD$.
$AC = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-4)^2} = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.
$BD = \sqrt{(2-1)^2 + (1-0)^2 + (8-(-1))^2} = \sqrt{1^2 + 1^2 + 9^2} = \sqrt{1+1+81} = \sqrt{83}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,the points $A, B, C$,and $D$ form a parallelogram.

(A) Let $A = (1,2,3)$,$B = (3,-1,5)$,and $C = (4,0,-3)$.
First,we find the direction ratios of the sides:
Direction ratios of $\overline{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$.
Direction ratios of $\overline{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$.
Check for orthogonality: $\vec{AB} \cdot \vec{AC} = (2)(3) + (-3)(-2) + (2)(-6) = 6 + 6 - 12 = 0$.
Since the dot product is $0$,$\overline{AB} \perp \overline{AC}$,which means $\angle A = 90^{\circ}$.
In a right-angled triangle,the orthocentre $H$ is the vertex where the right angle is located. Thus,$H = A = (1, 2, 3)$.
The circumcentre $S$ of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.
$S = \left( \frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2} \right) = \left( \frac{7}{2}, -\frac{1}{2}, 1 \right)$.
The distance $HS$ is given by the distance formula:
$HS = \sqrt{\left( \frac{7}{2} - 1 \right)^2 + \left( -\frac{1}{2} - 2 \right)^2 + (1 - 3)^2}$
$HS = \sqrt{\left( \frac{5}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-2)^2}$
$HS = \sqrt{\frac{25}{4} + \frac{25}{4} + 4} = \sqrt{\frac{50}{4} + \frac{16}{4}} = \sqrt{\frac{66}{4}} = \sqrt{\frac{33}{2}}$.
Thus,the correct option is $(A)$.

(B) Total ways to divide $75$ students into two sections of $30$ and $45$ is $^{75}C_{30}$.
Let the two particular students be $S_1$ and $S_2$.
Case $I$: Both $S_1$ and $S_2$ are in the section of $30$ students. The number of ways is $^{73}C_{28}$.
Case $II$: Both $S_1$ and $S_2$ are in the section of $45$ students. The number of ways is $^{73}C_{43}$.
The required probability is $\frac{^{73}C_{28} + ^{73}C_{43}}{^{75}C_{30}}$.
Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we get:
$\frac{\frac{73!}{28!45!} + \frac{73!}{43!30!}}{\frac{75!}{30!45!}} = \frac{73!}{75!} \times \left( \frac{30!45!}{28!45!} + \frac{30!45!}{43!30!} \right)$
$= \frac{1}{75 \times 74} \times (30 \times 29 + 45 \times 44)$
$= \frac{870 + 1980}{5550} = \frac{2850}{5550} = \frac{285}{555} = \frac{19}{37}$.

(A) Total number of students $= 50$. The total number of ways to divide $50$ students into two groups of $20$ and $30$ is given by ${}^{50}C_{20} \times {}^{30}C_{30} = {}^{50}C_{20}$.
If both $Ram$ and $Rahim$ are in the first group (size $20$),we need to choose $18$ more students from the remaining $48$ students. The number of ways is ${}^{48}C_{18}$.
If both $Ram$ and $Rahim$ are in the second group (size $30$),we need to choose $28$ more students from the remaining $48$ students. The number of ways is ${}^{48}C_{28}$.
The required probability is $P = \frac{{}^{48}C_{18} + {}^{48}C_{28}}{{}^{50}C_{20}}$.
Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$P = \frac{\frac{48!}{18!30!} + \frac{48!}{28!20!}}{\frac{50!}{20!30!}} = \frac{48!}{18!30!} \cdot \frac{20!30!}{50!} + \frac{48!}{28!20!} \cdot \frac{20!30!}{50!}$
$P = \frac{20 \times 19}{50 \times 49} + \frac{30 \times 29}{50 \times 49} = \frac{380 + 870}{2450} = \frac{1250}{2450} = \frac{25}{49}$.
Thus,the correct option is $A$.

(B) Let $X$ be the number drawn from box $A$ and $Y$ be the number drawn from box $B$. The total number of possible outcomes is $10 \times 10 = 100$.
We want to find the probability $P(X < Y)$.
The total outcomes can be divided into three cases: $X < Y$,$X > Y$,and $X = Y$.
The number of cases where $X = Y$ is $10$ (i.e.,$(1,1), (2,2), \ldots, (10,10)$).
Since the situation is symmetric,the number of cases where $X < Y$ is equal to the number of cases where $X > Y$.
Let $N$ be the number of cases where $X < Y$. Then $N + N + 10 = 100$.
$2N = 90 \implies N = 45$.
The required probability is $\frac{N}{100} = \frac{45}{100} = \frac{9}{20}$.
Hence,option $B$ is correct.

(D) Given $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$.
$e^y = \tan \left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$.
We know that $\tanh \left(\frac{y}{2}\right) = \frac{e^y - 1}{e^y + 1}$.
Substituting the value of $e^y$:
$\tanh \left(\frac{y}{2}\right) = \frac{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} - 1}{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} + 1} = \frac{(1 + \tan \frac{x}{2}) - (1 - \tan \frac{x}{2})}{(1 + \tan \frac{x}{2}) + (1 - \tan \frac{x}{2})} = \frac{2 \tan \frac{x}{2}}{2} = \tan \frac{x}{2}$.
Thus,option $D$ is correct.

(D) Given that,the circle $S_1 \equiv x^2+y^2+6x-2y+k=0$ and $S_2 \equiv x^2+y^2+2x-6y-15=0$.
Since $S_1$ bisects the circumference of $S_2$,the common chord of $S_1$ and $S_2$ must be the diameter of $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$.
$4x + 4y + k + 15 = 0$.
Since this chord is the diameter of $S_2$,it must pass through the center of $S_2$.
The center of $S_2$ is $(-g, -f) = (-1, 3)$.
Substituting $(-1, 3)$ into the chord equation:
$4(-1) + 4(3) + k + 15 = 0$.
$-4 + 12 + k + 15 = 0$.
$8 + k + 15 = 0$.
$k + 23 = 0$.
$k = -23$.

(D) The equation of the parabola is $y^2=4ax$,where $a=1$.
The normals to the parabola $y^2=4ax$ passing through a point $(h,k)$ are given by the cubic equation $my^3 + (2a-h)m^2 + k^2m - k = 0$.
For the point $(5,0)$,we have $h=5$ and $k=0$.
The equation becomes $m(y^2 + (2-5)) = 0$,which simplifies to $m(y^2-3)=0$.
The feet of the normals $(x_i, y_i)$ for $i=1, 2, 3$ are the points on the parabola.
The centroid $(G_x, G_y)$ of the triangle formed by the feet of the normals $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $G_x = \frac{x_1+x_2+x_3}{3}$ and $G_y = \frac{y_1+y_2+y_3}{3}$.
For a parabola $y^2=4ax$,the centroid of the triangle formed by the feet of the three normals passing through $(h,k)$ is $\left(\frac{2}{3}(h-2a), 0\right)$.
Substituting $h=5$ and $a=1$,we get the centroid as $\left(\frac{2}{3}(5-2(1)), 0\right) = \left(\frac{2}{3}(3), 0\right) = (2,0)$.

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given the midpoints of $AB, BC, CA$ are $M_1(3,0,0)$,$M_2(0,4,0)$,and $M_3(0,0,5)$ respectively.
Using the midpoint formula:
$x_1+x_2=6, x_2+x_3=0, x_3+x_1=0$
Solving these,we get $x_1=3, x_2=3, x_3=-3$.
Similarly for $y$ coordinates:
$y_1+y_2=0, y_2+y_3=8, y_3+y_1=0$
Solving these,we get $y_1=-4, y_2=4, y_3=4$.
Similarly for $z$ coordinates:
$z_1+z_2=0, z_2+z_3=0, z_3+z_1=10$
Solving these,we get $z_1=5, z_2=-5, z_3=5$.
Thus,the vertices are $A(3, -4, 5)$,$B(3, 4, -5)$,and $C(-3, 4, 5)$.
Now,calculate the squares of the side lengths:
$AB^2 = (3-3)^2 + (4-(-4))^2 + (-5-5)^2 = 0^2 + 8^2 + (-10)^2 = 64 + 100 = 164$.
$BC^2 = (-3-3)^2 + (4-4)^2 + (5-(-5))^2 = (-6)^2 + 0^2 + 10^2 = 36 + 100 = 136$.
$CA^2 = (3-(-3))^2 + (-4-4)^2 + (5-5)^2 = 6^2 + (-8)^2 + 0^2 = 36 + 64 = 100$.
Finally,$AB^2+BC^2+CA^2 = 164 + 136 + 100 = 400$.