System of co-ordinates, Direction cosines and direction ratios, Projection Questions in English

(D) Let the angles made by the line with $X, Y,$ and $Z-$ axes be $\alpha, \beta,$ and $\gamma$ respectively.
Given $\alpha = 30^o$ and $\beta = 45^o$.
The direction cosines of the line are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $\cos^2 30^o + \cos^2 45^o + \cos^2 \gamma = 1$.
$(\frac{\sqrt{3}}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{3}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.
$\frac{3+2}{4} + \cos^2 \gamma = 1 \implies \frac{5}{4} + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 1 - \frac{5}{4} = -\frac{1}{4}$.
Since $\cos^2 \gamma$ cannot be negative,such a line does not exist in real space.
Therefore,the correct option is $D$.

(D) We know that for direction cosines,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Since $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) = 3 - 1 = 2$.
The given equation is $2(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 3 \sec^2 \frac{\theta}{2}$.
Substituting the value,$2(2) = 3 \sec^2 \frac{\theta}{2} \Rightarrow 4 = 3 \sec^2 \frac{\theta}{2}$.
$\sec^2 \frac{\theta}{2} = \frac{4}{3} \Rightarrow \cos^2 \frac{\theta}{2} = \frac{3}{4}$.
Using the identity $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,we get $\cos \theta = 2(\frac{3}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.
Thus,$\theta = \frac{\pi}{3}$.

(B) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$.
The direction ratios $(DRs)$ of the median $AD$ are $\left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal. Thus,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.
From $\frac{\lambda - 5}{2} = 1$,we get $\lambda - 5 = 2$,so $\lambda = 7$.
From $\frac{\mu - 8}{2} = 1$,we get $\mu - 8 = 2$,so $\mu = 10$.
Now,calculate the value of $(\lambda^3 + \mu^3 + 5) = 7^3 + 10^3 + 5 = 343 + 1000 + 5 = 1348$.

(C) Let the direction cosines of the line be $l, m, n$. Since the line makes an angle $\theta$ with the $x$ and $y$ axes,we have $l = \cos \theta$ and $m = \cos \theta$.
Let the angle with the $z$-axis be $\phi$. Then $n = \cos \phi$.
The condition for direction cosines is $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \theta + \cos^2 \theta + \cos^2 \phi = 1$,which implies $2 \cos^2 \theta + \cos^2 \phi = 1$.
Thus,$\cos^2 \phi = 1 - 2 \cos^2 \theta = - \cos 2 \theta$.
Since $\cos^2 \phi \ge 0$,we must have $-\cos 2 \theta \ge 0$,which implies $\cos 2 \theta \le 0$.
Given $0 < \theta \le \frac{\pi}{2}$,we have $0 < 2 \theta \le \pi$.
For $\cos 2 \theta \le 0$ in the interval $(0, \pi]$,we have $\frac{\pi}{2} \le 2 \theta \le \pi$.
Dividing by $2$,we get $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$.
Therefore,the set of all values of $\theta$ is $\left[ \frac{\pi}{4}, \frac{\pi}{2} \right]$.

(C) Given the equations for direction cosines: $l+m+n=0$ and $l^2+m^2-n^2=0$.
From the first equation,$l+m = -n$. Squaring both sides,we get $l^2+m^2+2lm = n^2$.
Substituting $l^2+m^2 = n^2$ from the second equation into this,we get $n^2+2lm = n^2$,which implies $2lm = 0$,so $lm = 0$.
This means either $l=0$ or $m=0$.
Case $1$: If $l=0$,then $m+n=0 \Rightarrow m=-n$. Since $l^2+m^2+n^2=1$,we have $0^2+(-n)^2+n^2=1 \Rightarrow 2n^2=1 \Rightarrow n = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ or $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $l+n=0 \Rightarrow l=-n$. Similarly,$l^2+0^2+n^2=1 \Rightarrow 2n^2=1 \Rightarrow n = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$ or $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
Let the two lines have direction cosines $\vec{u_1} = (0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $\vec{u_2} = (-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
The cosine of the angle $\theta$ between them is $|\vec{u_1} \cdot \vec{u_2}| = |(0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0) + (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,the acute angle $\theta = 60^o$.

(B) Let the projections of the line segment on the $x, y$ and $z-$ axes be $p_x = 2$,$p_y = 3$,and $p_z = 6$ respectively.
The length of the line segment $L$ in $3-$ dimensional space is given by the formula $L = \sqrt{p_x^2 + p_y^2 + p_z^2}$.
Substituting the given values,we get $L = \sqrt{2^2 + 3^2 + 6^2}$.
$L = \sqrt{4 + 9 + 36}$.
$L = \sqrt{49}$.
$L = 7$.
Therefore,the length of the line segment is $7$.

(C) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:
$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$
The direction ratios of the median $AD$ are:
$a = \frac{\lambda - 1}{2} - 2 = \frac{\lambda - 5}{2}$
$b = 4 - 3 = 1$
$c = \frac{\mu + 2}{2} - 5 = \frac{\mu - 8}{2}$
Since the median $AD$ is equally inclined with the axes,its direction cosines $l, m, n$ are equal,i.e.,$|l| = |m| = |n|$. Since $l^2 + m^2 + n^2 = 1$,we have $l = m = n = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction ratios $a, b, c$ must be proportional to $1, 1, 1$. Since $b = 1$,we must have $a = 1$ and $c = 1$.
Setting $a = 1$:
$\frac{\lambda - 5}{2} = 1 \Rightarrow \lambda - 5 = 2 \Rightarrow \lambda = 7$
Setting $c = 1$:
$\frac{\mu - 8}{2} = 1 \Rightarrow \mu - 8 = 2 \Rightarrow \mu = 10$
Therefore,$(\lambda, \mu) = (7, 10)$.

(C) Let the direction angles of the unit vector $\vec{r}$ be $\alpha = \frac{\pi}{3}$,$\beta = \frac{\pi}{4}$,and $\gamma = \theta$.
The sum of the squares of the direction cosines of a unit vector is always $1$,given by the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values: $\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2 \theta = 1$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \theta = 1$.
$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$.
$\frac{3}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$.
Therefore,$\cos \theta = \pm \frac{1}{2}$.
Given $\theta \in (0, \pi)$,if $\cos \theta = \frac{1}{2}$,then $\theta = \frac{\pi}{3}$. If $\cos \theta = -\frac{1}{2}$,then $\theta = \frac{2\pi}{3}$.
Comparing with the options,$\frac{2\pi}{3}$ is the correct value.

(A) The direction ratios of a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ are the components $(x, y, z)$.
For the given vector $\vec{a} = 1\hat{i} + 1\hat{j} - 2\hat{k}$,the direction ratios are $a = 1, b = 1, c = -2$.
Next,we calculate the magnitude of the vector: $|\vec{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
The direction cosines $(l, m, n)$ are given by $l = \frac{a}{|\vec{a}|}, m = \frac{b}{|\vec{a}|}, n = \frac{c}{|\vec{a}|}$.
Substituting the values,we get $l = \frac{1}{\sqrt{6}}, m = \frac{1}{\sqrt{6}}, n = \frac{-2}{\sqrt{6}}$.
Thus,the direction cosines are $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}\right)$.

(A) The given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.
The vector $\overrightarrow{AB}$ is given by $(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
$\overrightarrow{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
The direction cosines are given by $\left(\frac{a}{|\overrightarrow{AB}|}, \frac{b}{|\overrightarrow{AB}|}, \frac{c}{|\overrightarrow{AB}|}\right)$,where $\overrightarrow{AB} = a\hat{i} + b\hat{j} + c\hat{k}$.
Thus,the direction cosines are $\left(-\frac{2}{6}, -\frac{4}{6}, \frac{4}{6}\right) = \left(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)$.

(A) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$,where $\alpha, \beta, \gamma$ are the angles made by the line with the positive directions of the $x, y,$ and $z$-axes respectively.
Given $\alpha = 90^{\circ}, \beta = 60^{\circ}, \gamma = 30^{\circ}$.
Therefore,$l = \cos 90^{\circ} = 0$.
$m = \cos 60^{\circ} = \frac{1}{2}$.
$n = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
Thus,the direction cosines are $0, \frac{1}{2}, \frac{\sqrt{3}}{2}$.

(A) Let the direction ratios be $a = 2, b = -1, c = -2$.
The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$.
Substituting the values,we get $l = \frac{2}{3}, m = -\frac{1}{3}, n = -\frac{2}{3}$.
Thus,the direction cosines are $\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}$.

(A) The direction cosines of a line passing through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are given by $\frac{x_2-x_1}{PQ}, \frac{y_2-y_1}{PQ}, \frac{z_2-z_1}{PQ}$,where $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
Given points are $P(-2, 4, -5)$ and $Q(1, 2, 3)$.
First,calculate the distance $PQ$:
$PQ = \sqrt{(1 - (-2))^2 + (2 - 4)^2 + (3 - (-5))^2} = \sqrt{(3)^2 + (-2)^2 + (8)^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$.
Now,the direction cosines are:
$l = \frac{1 - (-2)}{\sqrt{77}} = \frac{3}{\sqrt{77}}$
$m = \frac{2 - 4}{\sqrt{77}} = \frac{-2}{\sqrt{77}}$
$n = \frac{3 - (-5)}{\sqrt{77}} = \frac{8}{\sqrt{77}}$
Thus,the direction cosines are $\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$.

(A) The $x$-axis makes angles $0^{\circ}, 90^{\circ}$,and $90^{\circ}$ with the $x$,$y$,and $z$-axes respectively.
Therefore,the direction cosines of the $x$-axis are $\cos 0^{\circ}, \cos 90^{\circ}, \cos 90^{\circ}$,which are $(1, 0, 0)$.
Similarly,the $y$-axis makes angles $90^{\circ}, 0^{\circ}$,and $90^{\circ}$ with the $x$,$y$,and $z$-axes respectively.
Therefore,the direction cosines of the $y$-axis are $\cos 90^{\circ}, \cos 0^{\circ}, \cos 90^{\circ}$,which are $(0, 1, 0)$.
Finally,the $z$-axis makes angles $90^{\circ}, 90^{\circ}$,and $0^{\circ}$ with the $x$,$y$,and $z$-axes respectively.
Therefore,the direction cosines of the $z$-axis are $\cos 90^{\circ}, \cos 90^{\circ}, \cos 0^{\circ}$,which are $(0, 0, 1)$.

(A) Let the direction angles of the line be $\alpha = 90^{\circ}, \beta = 135^{\circ},$ and $\gamma = 45^{\circ}$.
The direction cosines $(l, m, n)$ of a line are given by $l = \cos \alpha, m = \cos \beta,$ and $n = \cos \gamma$.
Calculating each value:
$l = \cos 90^{\circ} = 0$
$m = \cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}}$
$n = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
Therefore,the direction cosines of the line are $0, -\frac{1}{\sqrt{2}},$ and $\frac{1}{\sqrt{2}}$.

(A) Let the direction cosines of the line be $l, m, n$. Since the line makes equal angles $\alpha$ with the coordinate axes,we have $l = \cos \alpha$,$m = \cos \alpha$,and $n = \cos \alpha$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
This simplifies to $3 \cos^2 \alpha = 1$,which implies $\cos^2 \alpha = \frac{1}{3}$.
Therefore,$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction cosines are $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$.

(A) Given direction ratios are $a = -18, b = 12, c = -4.$
The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}.$
$= \sqrt{324 + 144 + 16} = \sqrt{484} = 22.$
The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}.$
$l = \frac{-18}{22} = \frac{-9}{11},$
$m = \frac{12}{22} = \frac{6}{11},$
$n = \frac{-4}{22} = \frac{-2}{11}.$
Thus,the direction cosines are $\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}.$

Let the vertices of $\triangle ABC$ be $A(3,5,-4), B(-1,1,2),$ and $C(-5,-5,-2)$.
The direction ratios of side $AB$ are $(-1-3), (1-5), (2-(-4)),$ which are $-4, -4, 6$.
The magnitude of vector $\vec{AB}$ is $\sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$.
Thus,the direction cosines of $AB$ are $\frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}},$ which simplify to $-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$.
The direction ratios of side $BC$ are $(-5-(-1)), (-5-1), (-2-2),$ which are $-4, -6, -4$.
The magnitude of vector $\vec{BC}$ is $\sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$.
Thus,the direction cosines of $BC$ are $\frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}},$ which simplify to $-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}}$.
The direction ratios of side $CA$ are $(3-(-5)), (5-(-5)), (-4-(-2)),$ which are $8, 10, -2$.
The magnitude of vector $\vec{CA}$ is $\sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$.
Thus,the direction cosines of $CA$ are $\frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}},$ which simplify to $\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}}$.

Two lines with direction cosines $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are perpendicular to each other if $l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0$.
$(i)$ For the lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{12}{13}\right)\left(\frac{4}{13}\right) + \left(\frac{-3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$
$= \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0$.
Thus,these two lines are perpendicular.
$(ii)$ For the lines with direction cosines $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ and $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{4}{13}\right)\left(\frac{3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right) + \left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$
$= \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0$.
Thus,these two lines are perpendicular.
$(iii)$ For the lines with direction cosines $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ and $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{-3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right)$
$= \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0$.
Thus,these two lines are perpendicular.
Since all pairs are perpendicular,the three lines are mutually perpendicular.

(N/A) cube is a rectangular parallelepiped having equal length,breadth,and height.
Let the vertices of the cube be defined such that the diagonals are $OE, AF, BG,$ and $CD$.
Let the side length of the cube be $a$.
The coordinates of the vertices are $O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), D(a,a,0), E(a,a,a), F(0,a,a), G(a,0,a)$.
The direction vectors of the four diagonals are:
$d_1 = (a, a, a) \implies \text{unit vector } \hat{d}_1 = \frac{1}{\sqrt{3}}(1, 1, 1)$
$d_2 = (-a, a, a) \implies \text{unit vector } \hat{d}_2 = \frac{1}{\sqrt{3}}(-1, 1, 1)$
$d_3 = (a, -a, a) \implies \text{unit vector } \hat{d}_3 = \frac{1}{\sqrt{3}}(1, -1, 1)$
$d_4 = (a, a, -a) \implies \text{unit vector } \hat{d}_4 = \frac{1}{\sqrt{3}}(1, 1, -1)$
Let the direction cosines of the given line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.
The cosine of the angle between the line and a diagonal with unit vector $\hat{d}$ is given by the dot product: $\cos \theta = |l \cdot d_x + m \cdot d_y + n \cdot d_z|$.
Thus,$\cos \alpha = \frac{1}{\sqrt{3}}(l+m+n)$,$\cos \beta = \frac{1}{\sqrt{3}}(-l+m+n)$,$\cos \gamma = \frac{1}{\sqrt{3}}(l-m+n)$,and $\cos \delta = \frac{1}{\sqrt{3}}(l+m-n)$.
Squaring and adding these values:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} [(l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2]$
$= \frac{1}{3} [ (l^2+m^2+n^2 + 2lm + 2mn + 2nl) + (l^2+m^2+n^2 + 2lm - 2mn - 2nl) + (l^2+m^2+n^2 - 2lm - 2mn + 2nl) + (l^2+m^2+n^2 - 2lm + 2mn - 2nl) ]$
$= \frac{1}{3} [ 4(l^2+m^2+n^2) ]$
Since $l^2+m^2+n^2 = 1$,the sum is $\frac{4}{3}(1) = \frac{4}{3}$.

(N/A) It is given that $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,
$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0$ ........$(1)$
$l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1$ ..........$(2)$
$l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=1$ ...........$(3)$
Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$.
$\therefore l l_{1} + m m_{1} + n n_{1} = 0$
$l l_{2} + m m_{2} + n n_{2} = 0$
$\therefore \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}}$
$\Rightarrow \frac{l^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2}} = \frac{m^{2}}{(n_{1} l_{2} - n_{2} l_{1})^{2}} = \frac{n^{2}}{(l_{1} m_{2} - l_{2} m_{1})^{2}}$
$= \frac{l^{2} + m^{2} + n^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}}$ .........$(4)$
Since $l, m, n$ are the direction cosines of the line,$l^{2} + m^{2} + n^{2} = 1$ ........$(5)$
Using the Lagrange identity:
$(l_{1}^{2} + m_{1}^{2} + n_{1}^{2})(l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})^{2} = (m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}$
From $(1), (2),$ and $(3),$ we obtain:
$1 \cdot 1 - 0^{2} = (m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2} = 1$ .........$(6)$
Substituting $(5)$ and $(6)$ in $(4),$ we get:
$\frac{l^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2}} = \frac{m^{2}}{(n_{1} l_{2} - n_{2} l_{1})^{2}} = \frac{n^{2}}{(l_{1} m_{2} - l_{2} m_{1})^{2}} = 1$
Thus,$l = m_{1} n_{2} - m_{2} n_{1}, m = n_{1} l_{2} - n_{2} l_{1}, n = l_{1} m_{2} - l_{2} m_{1}$.

(A) The octants are determined by the signs of the coordinates $(x, y, z)$.
For the point $(-3, 1, 2)$,the signs are $(-, +, +)$,which corresponds to the second octant.
For the point $(-3, 1, -2)$,the signs are $(-, +, -)$,which corresponds to the sixth octant.
Therefore,the points lie in the second and sixth octants respectively.

(A) In a three-dimensional Cartesian coordinate system,any point lying in the $XZ-$ plane has its $y-$ coordinate equal to $0$.
Therefore,the $y-$ coordinate of the point is $0$.

The octants are determined by the signs of the coordinates $(x, y, z)$:
$1$. $(1, 2, 3)$: All positive,so it lies in octant $I$.
$2$. $(4, -2, 3)$: $(+, -, +)$,so it lies in octant $IV$.
$3$. $(4, -2, -5)$: $(+, -, -)$,so it lies in octant $VIII$.
$4$. $(4, 2, -5)$: $(+, +, -)$,so it lies in octant $V$.
$5$. $(-4, 2, -5)$: $(-, +, -)$,so it lies in octant $VI$.
$6$. $(-4, 2, 5)$: $(-, +, +)$,so it lies in octant $II$.
$7$. $(-3, -1, 6)$: $(-, -, +)$,so it lies in octant $III$.
$8$. $(-2, -4, -7)$: $(-, -, -)$,so it lies in octant $VII$.

(N/A) In a three-dimensional coordinate system,any point in the $XY$-plane has its $z$-coordinate equal to $0$. Therefore,the coordinates of points in the $XY$-plane are of the form $(x, y, 0)$.

(C) The three coordinate planes ($XY$,$YZ$,and $ZX$) divide the three-dimensional space into $8$ octants.

(A) The direction cosines $(l, m, n)$ are given by the formulas:
$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$
Here,the direction ratios are $a = 1, b = 1, c = 2$.
First,calculate the magnitude:
$\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Substituting these values,we get:
$l = \frac{1}{\sqrt{6}}, m = \frac{1}{\sqrt{6}}, n = \frac{2}{\sqrt{6}}$.
Since direction cosines can be positive or negative depending on the direction of the line,the direction cosines are $\pm(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}})$.

(C) The direction ratios of the line passing through $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ are given by $(x_{2}-x_{1}, y_{2}-y_{1}, z_{2}-z_{1})$.
For points $P(2, 3, 5)$ and $Q(-1, 2, 4)$,the direction ratios are $(-1-2, 2-3, 4-5) = (-3, -1, -1)$.
The distance $PQ$ is $\sqrt{(-3)^{2} + (-1)^{2} + (-1)^{2}} = \sqrt{9 + 1 + 1} = \sqrt{11}$.
The direction cosines are obtained by dividing the direction ratios by the distance $PQ$.
Thus,the direction cosines are $\left(\frac{-3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}\right)$ or $\left(\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right)$.

(A) The direction cosines of a line that makes angles $\alpha, \beta, \gamma$ with the positive $x, y, z$-axes are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
Given $\alpha = 30^{\circ}$,$\beta = 60^{\circ}$,and $\gamma = 90^{\circ}$.
Thus,the direction cosines are:
$l = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$
$m = \cos 60^{\circ} = \frac{1}{2}$
$n = \cos 90^{\circ} = 0$
Therefore,the direction cosines are $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right)$.

(A) Let the direction angles of $\overrightarrow{OA}$ be $\alpha = 60^{\circ}$,$\beta = 45^{\circ}$,and $\gamma$.
We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.
Substituting the values,we get $\cos^{2} 60^{\circ} + \cos^{2} 45^{\circ} + \cos^{2} \gamma = 1$.
$\Rightarrow (\frac{1}{2})^{2} + (\frac{1}{\sqrt{2}})^{2} + \cos^{2} \gamma = 1$.
$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^{2} \gamma = 1$.
$\Rightarrow \frac{3}{4} + \cos^{2} \gamma = 1$.
$\Rightarrow \cos^{2} \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.
$\Rightarrow \cos \gamma = \pm \frac{1}{2}$.
Assuming $\gamma$ is acute,$\cos \gamma = \frac{1}{2}$.
The direction cosines are $l = \cos 60^{\circ} = \frac{1}{2}$,$m = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$n = \cos 60^{\circ} = \frac{1}{2}$.
The position vector $\overrightarrow{OA} = |\overrightarrow{OA}| (l\hat{i} + m\hat{j} + n\hat{k})$.
$\overrightarrow{OA} = 10 (\frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k})$.
$\overrightarrow{OA} = 5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}$.

(A) Given equations are $l+m+n=0$ $(1)$ and $l^{2}+m^{2}-n^{2}=0$ $(2)$.
From $(1)$,$n = -(l+m)$. Substituting this into $(2)$:
$l^{2}+m^{2}-(-(l+m))^{2}=0$
$l^{2}+m^{2}-(l^{2}+m^{2}+2lm)=0$
$-2lm=0 \Rightarrow lm=0$.
Case $1$: $l=0$. From $(1)$,$m+n=0 \Rightarrow m=-n$. Direction ratios are $(0, -n, n)$ or $(0, -1, 1)$.
Case $2$: $m=0$. From $(1)$,$l+n=0 \Rightarrow l=-n$. Direction ratios are $(-n, 0, n)$ or $(-1, 0, 1)$.
Let $\vec{a} = 0\hat{i} - 1\hat{j} + 1\hat{k}$ and $\vec{b} = -1\hat{i} + 0\hat{j} + 1\hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(0)(-1) + (-1)(0) + (1)(1)|}{\sqrt{0^2+(-1)^2+1^2} \sqrt{(-1)^2+0^2+1^2}} = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(A) We have $l, m, n$ and $l+\delta l, m+\delta m, n+\delta n$ as direction cosines of a variable line in two different positions.
$\therefore l^{2}+m^{2}+n^{2}=1 \dots (i)$
And $(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1 \dots (ii)$
$\Rightarrow l^{2}+m^{2}+n^{2}+\delta l^{2}+\delta m^{2}+\delta n^{2}+2(l \delta l+m \delta m+n \delta n)=1$
$\Rightarrow \delta l^{2}+\delta m^{2}+\delta n^{2}=-2(l \delta l+m \delta m+n \delta n) \left[\because l^{2}+m^{2}+n^{2}=1\right]$
$\Rightarrow l \delta l+m \delta m+n \delta n=-\frac{1}{2}(\delta l^{2}+\delta m^{2}+\delta n^{2}) \dots (iii)$
Now,$\vec{a}$ and $\vec{b}$ are unit vectors along a line with direction cosines $l, m, n$ and $(l+\delta l), (m+\delta m), (n+\delta n)$ respectively.
$\therefore \vec{a}=l \hat{i}+m \hat{j}+n \hat{k}$ and $\vec{b}=(l+\delta l) \hat{i}+(m+\delta m) \hat{j}+(n+\delta n) \hat{k}$
$\Rightarrow \cos \delta \theta = \vec{a} \cdot \vec{b} = l(l+\delta l)+m(m+\delta m)+n(n+\delta n)$
$= (l^{2}+m^{2}+n^{2})+(l \delta l+m \delta m+n \delta n)$
$= 1-\frac{1}{2}(\delta l^{2}+\delta m^{2}+\delta n^{2}) \text{ [using Eq. } (iii)]$
$\Rightarrow 2(1-\cos \delta \theta) = \delta l^{2}+\delta m^{2}+\delta n^{2}$
$\Rightarrow 2(2 \sin^{2} \frac{\delta \theta}{2}) = \delta l^{2}+\delta m^{2}+\delta n^{2} \left[\because 1-\cos \theta = 2 \sin^{2} \frac{\theta}{2}\right]$
$\Rightarrow 4(\frac{\delta \theta}{2})^{2} = \delta l^{2}+\delta m^{2}+\delta n^{2} \left[\text{since } \delta \theta \text{ is small, } \sin \frac{\delta \theta}{2} \approx \frac{\delta \theta}{2}\right]$
$\therefore \delta \theta^{2} = \delta l^{2}+\delta m^{2}+\delta n^{2}$

Let a vector be equally inclined to the axes $OX, OY$,and $OZ$ at an angle $\alpha.$
Then,the direction cosines of the vector are $l = \cos \alpha, m = \cos \alpha,$ and $n = \cos \alpha.$
We know that for any vector,the sum of the squares of its direction cosines is $1.$
Therefore,$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1.$
This simplifies to $3 \cos^2 \alpha = 1.$
Solving for $\cos \alpha,$ we get $\cos^2 \alpha = \frac{1}{3},$ which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}.$
Assuming the vector is in the first octant,the direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right).$

(N/A) Let the direction cosines of the vector $\vec{r}$ be $l, m, n$.
Given that the vector makes an angle of $\frac{\pi}{4}$ with the $y$-axis and $\frac{\pi}{2}$ with the $z$-axis.
Thus,$m = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $n = \cos(\frac{\pi}{2}) = 0$.
We know that for any vector,$l^2 + m^2 + n^2 = 1$.
Substituting the values,$l^2 + (\frac{1}{\sqrt{2}})^2 + 0^2 = 1$.
$l^2 + \frac{1}{2} = 1 \Rightarrow l^2 = \frac{1}{2} \Rightarrow l = \pm \frac{1}{\sqrt{2}}$.
The unit vector $\hat{r}$ is given by $l\hat{i} + m\hat{j} + n\hat{k} = \pm \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$.
The required vector $\vec{r}$ is given by $|\vec{r}| \hat{r} = 3\sqrt{2} (\pm \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k})$.
Therefore,$\vec{r} = \pm 3\hat{i} + 3\hat{j}$.

(NONE) Given equations are $l+m-n=0$ and $l^{2}+m^{2}-n^{2}=0$.
From the first equation,$n = l+m$.
Substituting this into the second equation: $l^{2}+m^{2}=(l+m)^{2} = l^{2}+m^{2}+2lm$.
This simplifies to $2lm = 0$,which means $l=0$ or $m=0$.
Case $1$: If $l=0$,then $n=m$. Since $l^{2}+m^{2}+n^{2}=1$,we have $0^{2}+m^{2}+m^{2}=1$,so $2m^{2}=1$,$m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $n=l$. Since $l^{2}+m^{2}+n^{2}=1$,we have $l^{2}+0^{2}+l^{2}=1$,so $2l^{2}=1$,$l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Let the direction cosines be $\vec{u} = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $\vec{v} = (\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
The cosine of the angle $\alpha$ between them is $\cos \alpha = |\vec{u} \cdot \vec{v}| = |0 \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot 0 + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}| = \frac{1}{2}$.
We need to find $\sin^{4} \alpha + \cos^{4} \alpha$.
Since $\cos \alpha = \frac{1}{2}$,$\sin^{2} \alpha = 1 - \cos^{2} \alpha = 1 - \frac{1}{4} = \frac{3}{4}$.
Then $\sin^{4} \alpha + \cos^{4} \alpha = (\frac{3}{4})^{2} + (\frac{1}{2})^{2} = \frac{9}{16} + \frac{1}{4} = \frac{9+4}{16} = \frac{13}{16}$.

(A) Given equations are $n = 2(l + m)$ and $mn + nl + lm = 0$.
Substituting $n = 2(l + m)$ into the second equation:
$m(2l + 2m) + 2l(l + m) + lm = 0$
$2lm + 2m^2 + 2l^2 + 2lm + lm = 0$
$2l^2 + 5lm + 2m^2 = 0$
Dividing by $m^2$,we get $2t^2 + 5t + 2 = 0$ where $t = \frac{l}{m}$.
Solving the quadratic equation: $(2t + 1)(t + 2) = 0$,so $t = -\frac{1}{2}$ or $t = -2$.
Case $1$: If $\frac{l}{m} = -2$,then $l = -2m$. Substituting into $n = 2(l + m)$,we get $n = 2(-2m + m) = -2m$.
The direction ratios are $(-2m, m, -2m)$,which simplifies to $(-2, 1, -2)$.
Case $2$: If $\frac{l}{m} = -\frac{1}{2}$,then $m = -2l$. Substituting into $n = 2(l + m)$,we get $n = 2(l - 2l) = -2l$.
The direction ratios are $(l, -2l, -2l)$,which simplifies to $(1, -2, -2)$.
Let the direction ratios be $\vec{a} = (-2, 1, -2)$ and $\vec{b} = (1, -2, -2)$.
The cosine of the angle $\theta$ is given by $\cos \theta = \frac{|a_1b_1 + a_2b_2 + a_3b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$.
$\cos \theta = \frac{|(-2)(1) + (1)(-2) + (-2)(-2)|}{\sqrt{4+1+4} \sqrt{1+4+4}} = \frac{|-2 - 2 + 4|}{3 \times 3} = 0$.
Therefore,$\theta = \frac{\pi}{2}$.

(A) Given the relations for direction cosines $l, m, n$ are $l+m-n=0$ and $3l^{2}+m^{2}+cnl=0$.
From the first equation,we have $n = l+m$.
Substituting this into the second equation: $3l^{2}+m^{2}+cl(l+m)=0$.
Expanding this,we get $3l^{2}+m^{2}+cl^{2}+clm=0$.
Grouping the terms,we have $(3+c)l^{2}+clm+m^{2}=0$.
Dividing by $m^{2}$ (assuming $m \neq 0$),we get $(3+c)(\frac{l}{m})^{2}+c(\frac{l}{m})+1=0$.
Since the two lines are parallel,the roots of this quadratic equation in $(\frac{l}{m})$ must be equal.
Therefore,the discriminant $D = b^{2}-4ac = 0$.
$c^{2}-4(3+c)(1) = 0$.
$c^{2}-4c-12=0$.
Factoring the quadratic,we get $(c-6)(c+2)=0$.
This gives $c=6$ or $c=-2$.
Since we need the positive value of $c$,we have $c=6$.

(A) Given $P(x, y, z)$ in the first octant,$OP = \gamma = \sqrt{x^2+y^2+z^2}$.
The projection of $P$ on the $xy$-plane is $Q(x, y, 0)$.
Let $OQ = r = \sqrt{x^2+y^2}$.
Given the angle between $OQ$ and the positive $x$-axis is $\theta$,we have $x = r \cos \theta$ and $y = r \sin \theta$.
Given the angle between $OP$ and the positive $z$-axis is $\phi$,we have $z = OP \cos \phi = \gamma \cos \phi$.
Also,$r = OP \sin \phi = \gamma \sin \phi$.
Thus,$x = \gamma \sin \phi \cos \theta$,$y = \gamma \sin \phi \sin \theta$,and $z = \gamma \cos \phi$.
The distance of $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2+z^2}$.
Substituting the values: $\sqrt{(\gamma \sin \phi \sin \theta)^2 + (\gamma \cos \phi)^2} = \gamma \sqrt{\sin^2 \phi \sin^2 \theta + \cos^2 \phi}$.
Using $\cos^2 \phi = 1 - \sin^2 \phi$,we get $\gamma \sqrt{\sin^2 \phi \sin^2 \theta + 1 - \sin^2 \phi} = \gamma \sqrt{1 - \sin^2 \phi (1 - \sin^2 \theta)} = \gamma \sqrt{1 - \sin^2 \phi \cos^2 \theta}$.

(A) Let $\alpha, \beta, \gamma$ be the angles made by the line with the positive $x, y, z$-axes respectively.
Given $\beta = \frac{\alpha}{2}$ and $\gamma = \frac{\alpha}{2}$.
The direction cosines of the line satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values: $\cos^2 \alpha + \cos^2(\frac{\alpha}{2}) + \cos^2(\frac{\alpha}{2}) = 1$.
$\cos^2 \alpha + 2 \cos^2(\frac{\alpha}{2}) = 1$.
Using the identity $\cos^2(\frac{\alpha}{2}) = \frac{1 + \cos \alpha}{2}$,we get $\cos^2 \alpha + 2(\frac{1 + \cos \alpha}{2}) = 1$.
$\cos^2 \alpha + 1 + \cos \alpha = 1$.
$\cos^2 \alpha + \cos \alpha = 0$.
$\cos \alpha(\cos \alpha + 1) = 0$.
This implies $\cos \alpha = 0$ or $\cos \alpha = -1$.
If $\cos \alpha = 0$,then $\alpha = \frac{\pi}{2}$,so $\beta = \frac{\pi}{4}$.
If $\cos \alpha = -1$,then $\alpha = \pi$,so $\beta = \frac{\pi}{2}$.
The possible values for $\beta$ are $\frac{\pi}{4}$ and $\frac{\pi}{2}$.
The sum of all possible values of $\beta$ is $\frac{\pi}{4} + \frac{\pi}{2} = \frac{3 \pi}{4}$.

(C) Let the angles made by the line with the positive $X$,$Y$,and $Z$ axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = 45^{\circ}$ and $\beta = \gamma$.
Using the direction cosine property $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:
$\cos^2 45^{\circ} + \cos^2 \beta + \cos^2 \beta = 1$
$\left(\frac{1}{\sqrt{2}}\right)^2 + 2\cos^2 \beta = 1$
$\frac{1}{2} + 2\cos^2 \beta = 1$
$2\cos^2 \beta = \frac{1}{2}$
$\cos^2 \beta = \frac{1}{4}$
$\cos \beta = \frac{1}{2}$ (since angles are acute)
$\beta = 60^{\circ}$.
Thus,$\beta = 60^{\circ}$ and $\gamma = 60^{\circ}$.
The sum of the angles is $\alpha + \beta + \gamma = 45^{\circ} + 60^{\circ} + 60^{\circ} = 165^{\circ}$.

(C) Let the direction angles be $\alpha, \beta, \gamma$ with $X, Y, Z$ axes respectively.
Given $\alpha = 120^{\circ}$ and $\gamma = 60^{\circ}$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2(120^{\circ}) + \cos^2 \beta + \cos^2(60^{\circ}) = 1$.
$(-\frac{1}{2})^2 + \cos^2 \beta + (\frac{1}{2})^2 = 1$.
$\frac{1}{4} + \cos^2 \beta + \frac{1}{4} = 1$.
$\cos^2 \beta = 1 - \frac{1}{2} = \frac{1}{2}$.
$\cos \beta = \pm \frac{1}{\sqrt{2}}$.
Thus,$\beta = 45^{\circ}$ or $\beta = 135^{\circ}$.
Since $135^{\circ}$ is one of the options,the correct answer is $135^{\circ}$.

(A) Let point $P(x, y, z)$ divide the line segment joining the points $A(5, -3, -2)$ and $B(1, 2, -2)$ in the ratio $m:1$.
Since the point $P$ lies on the $XOZ$-plane,its $y$-coordinate must be $0$.
Using the section formula,the $y$-coordinate of $P$ is given by $\frac{m(2) + 1(-3)}{m + 1} = 0$.
$\Rightarrow 2m - 3 = 0 \Rightarrow m = \frac{3}{2}$.
Now,we find the $x$ and $z$ coordinates using $m = \frac{3}{2}$:
$x = \frac{m(1) + 1(5)}{m + 1} = \frac{\frac{3}{2}(1) + 5}{\frac{3}{2} + 1} = \frac{\frac{3+10}{2}}{\frac{5}{2}} = \frac{13}{5}$.
$z = \frac{m(-2) + 1(-2)}{m + 1} = \frac{\frac{3}{2}(-2) - 2}{\frac{3}{2} + 1} = \frac{-3 - 2}{\frac{5}{2}} = \frac{-5}{\frac{5}{2}} = -2$.
Therefore,the required point is $\left(\frac{13}{5}, 0, -2\right)$.

(B) Let the direction angles of the line be $\alpha = 45^{\circ}$,$\beta = 60^{\circ}$,and $\gamma = \theta$.
We know that the sum of the squares of the direction cosines is $1$,i.e.,$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$.
Substituting the given values: $\cos^{2}(45^{\circ}) + \cos^{2}(60^{\circ}) + \cos^{2}\theta = 1$.
$(\frac{1}{\sqrt{2}})^{2} + (\frac{1}{2})^{2} + \cos^{2}\theta = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^{2}\theta = 1$.
$\frac{3}{4} + \cos^{2}\theta = 1$.
$\cos^{2}\theta = 1 - \frac{3}{4} = \frac{1}{4}$.
$\cos\theta = \pm \frac{1}{2}$.
Since the angle $\theta$ is obtuse,$\cos\theta$ must be negative,so $\cos\theta = -\frac{1}{2}$.
Therefore,$\theta = 120^{\circ}$.

(A) Let the points be $P(2,1,-3)$ and $Q(-1,0,2)$.
The vector $\vec{PQ} = (-1-2)\hat{i} + (0-1)\hat{j} + (2-(-3))\hat{k} = -3\hat{i} - 1\hat{j} + 5\hat{k}$.
The direction ratios of the line are $3, 2, 6$. The direction vector of the line is $\vec{v} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
The magnitude of the direction vector is $|\vec{v}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
The unit vector along the line is $\hat{u} = \frac{3\hat{i} + 2\hat{j} + 6\hat{k}}{7}$.
The projection of the line segment $PQ$ on the line is given by the absolute value of the dot product of $\vec{PQ}$ and $\hat{u}$:
Projection $= |\vec{PQ} \cdot \hat{u}| = |(-3\hat{i} - 1\hat{j} + 5\hat{k}) \cdot \frac{(3\hat{i} + 2\hat{j} + 6\hat{k})}{7}|$
$= |\frac{(-3)(3) + (-1)(2) + (5)(6)}{7}| = |\frac{-9 - 2 + 30}{7}| = |\frac{19}{7}| = \frac{19}{7}$ units.

(D) The angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Given direction cosines are:
Line $1: (\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2})$
Line $2: (\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2})$
Calculating the dot product:
$\cos \theta = |(\frac{-\sqrt{3}}{4})(\frac{-\sqrt{3}}{4}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{-\sqrt{3}}{2})(\frac{\sqrt{3}}{2})|$
$\cos \theta = |\frac{3}{16} + \frac{1}{16} - \frac{3}{4}|$
$\cos \theta = |\frac{4}{16} - \frac{12}{16}| = |\frac{-8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,the angle $\theta = 60^{\circ}$.
Thus,the correct option is $D$.

(D) Let the points be $P(2, -1, 0)$ and $Q(3, 2, -1)$.
The vector $\vec{PQ}$ is given by $(3-2)\hat{i} + (2-(-1))\hat{j} + (-1-0)\hat{k} = 1\hat{i} + 3\hat{j} - 1\hat{k}$.
The direction ratios of the line are $1, 2, 2$. Thus,the direction vector $\vec{v}$ is $1\hat{i} + 2\hat{j} + 2\hat{k}$.
The unit vector $\hat{u}$ along the line is $\frac{1\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{1\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{9}} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
The projection of $\vec{PQ}$ on the line is the dot product $\vec{PQ} \cdot \hat{u}$.
Projection $= (1\hat{i} + 3\hat{j} - 1\hat{k}) \cdot (\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k})$.
Projection $= (1 \times \frac{1}{3}) + (3 \times \frac{2}{3}) + (-1 \times \frac{2}{3}) = \frac{1}{3} + 2 - \frac{2}{3} = \frac{1-2}{3} + 2 = -\frac{1}{3} + 2 = \frac{5}{3}$.
Thus,the correct option is $D$.

(A) Given the equations:
$6mn - 2nl + 5lm = 0$ $(1)$
$3l + m + 5n = 0 \implies m = -(3l + 5n)$ $(2)$
Substitute $(2)$ into $(1)$:
$6n(-(3l + 5n)) - 2nl + 5l(-(3l + 5n)) = 0$
$-18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0$
$-15l^2 - 45ln - 30n^2 = 0$
Divide by $-15$:
$l^2 + 3ln + 2n^2 = 0$
$(l + n)(l + 2n) = 0$
Case $1$: $l = -n$. Then $m = -(3(-n) + 5n) = -2n$.
Direction ratios are $(-n, -2n, n)$,i.e.,$(1, 2, -1)$.
Case $2$: $l = -2n$. Then $m = -(3(-2n) + 5n) = n$.
Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.
Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} + \hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|-2 + 2 - 1|}{\sqrt{1+4+1} \sqrt{4+1+1}} = \frac{1}{\sqrt{6} \sqrt{6}} = \frac{1}{6}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - (\frac{1}{6})^2 = 1 - \frac{1}{36} = \frac{35}{36}$.
Therefore,$\sin \theta = \frac{\sqrt{35}}{6}$.

(B) Given equations are $\ell+m+n=0$ $(1)$ and $\ell^2+m^2-n^2=0$ $(2)$.
From $(1)$,$n = -(\ell+m)$.
Substituting this into $(2)$: $\ell^2+m^2-(-\ell-m)^2 = 0$.
$\ell^2+m^2-(\ell^2+m^2+2\ell m) = 0$.
$-2\ell m = 0$,which implies $\ell=0$ or $m=0$.
Case $1$: If $\ell=0$,then $n=-m$. The direction ratios are $(0, m, -m)$,which simplifies to $(0, 1, -1)$.
Case $2$: If $m=0$,then $n=-\ell$. The direction ratios are $(\ell, 0, -\ell)$,which simplifies to $(1, 0, -1)$.
Let the direction vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$.
$|\vec{a}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) Given equations are $l+m+n=0$ $(1)$ and $2mn+3ln-5lm=0$ $(2)$.
From $(1)$,$n = -(l+m)$.
Substituting this in $(2)$: $2m(-(l+m)) + 3l(-(l+m)) - 5lm = 0$.
$-2ml - 2m^2 - 3l^2 - 3lm - 5lm = 0$.
$-3l^2 - 10lm - 2m^2 = 0$,which implies $3l^2 + 10lm + 2m^2 = 0$.
Dividing by $m^2$,we get $3(l/m)^2 + 10(l/m) + 2 = 0$.
Let the direction ratios of the two lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
From the quadratic equation,$l_1/m_1 \times l_2/m_2 = 2/3$,so $l_1l_2/m_1m_2 = 2/3$,which means $3l_1l_2 = 2m_1m_2$.
Similarly,by eliminating $l$ or $m$,we can find the relations between direction cosines.
The angle $\theta$ between the lines is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
Using the property of lines satisfying these equations,the angle is $\frac{\pi}{3}$.

(A) Given relations are $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$.
From the first equation,$l = 5m - 3n$.
Substituting this into the second equation: $7(5m-3n)^2 + 5m^2 - 3n^2 = 0$.
$7(25m^2 - 30mn + 9n^2) + 5m^2 - 3n^2 = 0$.
$175m^2 - 210mn + 63n^2 + 5m^2 - 3n^2 = 0$.
$180m^2 - 210mn + 60n^2 = 0$.
Dividing by $30$,we get $6m^2 - 7mn + 2n^2 = 0$.
Factoring the quadratic: $(3m - 2n)(2m - n) = 0$.
Case $1$: $3m = 2n \Rightarrow m = \frac{2n}{3}$. Then $l = 5(\frac{2n}{3}) - 3n = \frac{10n-9n}{3} = \frac{n}{3}$.
Using $l^2 + m^2 + n^2 = 1$: $(\frac{n}{3})^2 + (\frac{2n}{3})^2 + n^2 = 1 \Rightarrow \frac{n^2}{9} + \frac{4n^2}{9} + n^2 = 1 \Rightarrow \frac{14n^2}{9} = 1 \Rightarrow n = \frac{3}{\sqrt{14}}$.
Then $l = \frac{1}{\sqrt{14}}$ and $m = \frac{2}{\sqrt{14}}$.
$l+m+n = \frac{1+2+3}{\sqrt{14}} = \frac{6}{\sqrt{14}}$.
Case $2$: $2m = n \Rightarrow m = \frac{n}{2}$. Then $l = 5(\frac{n}{2}) - 3n = \frac{5n-6n}{2} = -\frac{n}{2}$.
Using $l^2 + m^2 + n^2 = 1$: $(-\frac{n}{2})^2 + (\frac{n}{2})^2 + n^2 = 1 \Rightarrow \frac{n^2}{4} + \frac{n^2}{4} + n^2 = 1 \Rightarrow \frac{6n^2}{4} = 1 \Rightarrow n^2 = \frac{2}{3} \Rightarrow n = \sqrt{\frac{2}{3}}$.
Then $l = -\frac{1}{\sqrt{6}}$ and $m = \frac{1}{\sqrt{6}}$.
$l+m+n = \frac{-1+1+\sqrt{4}}{\sqrt{6}} = \frac{2}{\sqrt{6}}$.
Thus,the possible values are $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$.