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

(B) Given equations are $l+m+n=0$ and $2l^2+2m^2-n^2=0$.
From the first equation,$n = -(l+m)$.
Substituting this into the second equation:
$2l^2 + 2m^2 - (-(l+m))^2 = 0$
$2l^2 + 2m^2 - (l^2 + m^2 + 2lm) = 0$
$l^2 + m^2 - 2lm = 0$
$(l-m)^2 = 0 \Rightarrow l=m$.
If $l=m$,then $n = -(l+l) = -2l$.
The direction ratios of the lines are proportional to $(l, l, -2l)$,which simplifies to $(1, 1, -2)$.
Since the direction ratios are the same for both lines,the lines are parallel.
The angle $\theta$ between parallel lines is $0^{\circ}$ or $180^{\circ}$.
Given the options,$180^{\circ}$ is the correct choice.

(D) Let the direction cosines of the line be $l, m, n$. Since the line makes angles $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2}$ with the axes,we have $l = \cos(\frac{\alpha}{2}), m = \cos(\frac{\beta}{2}), n = \cos(\frac{\gamma}{2})$.
Since $l^2 + m^2 + n^2 = 1$,we have $\cos^2(\frac{\alpha}{2}) + \cos^2(\frac{\beta}{2}) + \cos^2(\frac{\gamma}{2}) = 1$.
We know that $\cos \theta = 2 \cos^2(\frac{\theta}{2}) - 1$,so $\cos^2(\frac{\theta}{2}) = \frac{1 + \cos \theta}{2}$.
Substituting this into the identity:
$\frac{1 + \cos \alpha}{2} + \frac{1 + \cos \beta}{2} + \frac{1 + \cos \gamma}{2} = 1$
$\frac{3 + \cos \alpha + \cos \beta + \cos \gamma}{2} = 1$
$3 + \cos \alpha + \cos \beta + \cos \gamma = 2$
$\cos \alpha + \cos \beta + \cos \gamma = 2 - 3 = -1$.

(B) Let the direction cosines of the line be $l, m, n$. Since the line makes the same angle $\alpha$ with the $x$ and $y$ axes,we have $l = \cos \alpha$ and $m = \cos \alpha$. Let the angle with the $z$-axis be $\theta$,so $n = \cos \theta$.
We know that $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \theta = 1$.
Using the identity $\cos^2 \phi = 1 - \sin^2 \phi$,we get $(1 - \sin^2 \alpha) + (1 - \sin^2 \alpha) + (1 - \sin^2 \theta) = 1$.
$3 - 2 \sin^2 \alpha - \sin^2 \theta = 1$.
Given $\sin^2 \theta = 2 \sin^2 \alpha$,substituting this into the equation:
$3 - 2 \sin^2 \alpha - 2 \sin^2 \alpha = 1$.
$3 - 4 \sin^2 \alpha = 1$.
$4 \sin^2 \alpha = 2$.
$\sin^2 \alpha = \frac{1}{2} = \sin^2 \frac{\pi}{4}$.
Thus,$\alpha = \frac{\pi}{4}$.

(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $\langle -1, 2, 2 \rangle$ and $\langle 0, 2, 1 \rangle$,the vector $\vec{n} = \langle a, b, c \rangle$ is parallel to the cross product of the two given vectors.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-1-0) + \hat{k}(-2-0) = -2\hat{i} + \hat{j} - 2\hat{k}$.
Thus,the direction ratios are $\langle -2, 1, -2 \rangle$ or $\langle 2, -1, 2 \rangle$.
The magnitude of the vector is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$.

(A) Given points are $O \equiv(0, 0, 0)$ and $P \equiv(1, 2, 3)$.
First,calculate the length of the vector $\overline{OP}$:
$|\overline{OP}| = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ of a vector $\overline{OP} = (x, y, z)$ are given by $\frac{x}{|\overline{OP}|}, \frac{y}{|\overline{OP}|}, \frac{z}{|\overline{OP}|}$.
Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.

(C) The given equation of the line is $\frac{x+2}{2} = \frac{2y-4}{3}$ and $z = -1$.
We can rewrite the equation as $\frac{x+2}{2} = \frac{2(y-2)}{3} = \frac{y-2}{3/2}$ and $z = -1$.
This represents a line in the $xy$-plane parallel to the $xy$-plane,where the $z$-coordinate is constant $(z = -1)$.
The direction ratios of the line are $(a, b, c) = (2, \frac{3}{2}, 0)$.
To find the direction cosines,we calculate the magnitude of the direction vector: $\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (\frac{3}{2})^2 + 0^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{16+9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
The direction cosines $(\ell, m, n)$ are given by $(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}})$.
Thus,$\ell = \pm \frac{2}{5/2} = \pm \frac{4}{5}$,$m = \pm \frac{3/2}{5/2} = \pm \frac{3}{5}$,and $n = \pm \frac{0}{5/2} = 0$.
Therefore,the direction cosines are $\pm \frac{4}{5}, \pm \frac{3}{5}, 0$.

(C) Let the direction cosines of the line be $l, m, n$. Since the line makes equal angles $\alpha$ with each of the coordinate axes,we have $l = \cos \alpha$,$m = \cos \alpha$,and $n = \cos \alpha$.
We know that for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{3}$.
Since the line is in the first octant $OXYZ$,the direction cosines must be positive,so $\cos \alpha = \frac{1}{\sqrt{3}}$.
Thus,$l = m = n = \frac{1}{\sqrt{3}}$.

(C) Let the direction angles of the line be $\alpha, \beta, \gamma$.
Since the line makes equal angles with the coordinate axes,we have $\alpha = \beta = \gamma$.
The direction cosines are given by $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.
We know the identity $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting $\alpha = \beta = \gamma$,we get $3 \cos^2 \alpha = 1$.
This gives $\cos^2 \alpha = \frac{1}{3}$,so $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Since the angles are given to be acute,$\cos \alpha, \cos \beta, \cos \gamma$ must be positive.
Therefore,the direction cosines are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

(B) Direction cosines are the cosines of the angles which a line makes with the positive coordinate axes. They are represented by $\langle l, m, n \rangle$ where $l, m, n$ correspond to the $x$-axis,$y$-axis,and $z$-axis respectively. The sum of the squares of the direction cosines is unity,i.e.,$l^{2} + m^{2} + n^{2} = 1$.
Since the line lies in the $ZOX$ plane,it is perpendicular to the $y$-axis. Therefore,the angle made with the $y$-axis is $90^{\circ}$,so $m = \cos(90^{\circ}) = 0$.
The line makes an angle of $30^{\circ}$ with the $Z$-axis,so $n = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}$.
Using the property $l^{2} + m^{2} + n^{2} = 1$,we substitute $m = 0$ and $n = \frac{\sqrt{3}}{2}$:
$l^{2} + 0^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2} = 1$
$l^{2} + \frac{3}{4} = 1$
$l^{2} = 1 - \frac{3}{4} = \frac{1}{4}$
$l = \pm \frac{1}{2}$.
Thus,the direction cosines are $\langle \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2} \rangle$.

(C) The formula for the cosine of the angle $\theta$ between two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Given $\theta = \frac{\pi}{3}$,$a_1=4, b_1=-3, c_1=5$ and $a_2=3, b_2=4, c_2=k$.
Substituting these values,we get $\cos \frac{\pi}{3} = \left| \frac{4(3) + (-3)(4) + 5k}{\sqrt{4^2 + (-3)^2 + 5^2} \sqrt{3^2 + 4^2 + k^2}} \right|$.
$\frac{1}{2} = \left| \frac{12 - 12 + 5k}{\sqrt{16 + 9 + 25} \sqrt{9 + 16 + k^2}} \right|$.
$\frac{1}{2} = \left| \frac{5k}{\sqrt{50} \sqrt{25 + k^2}} \right| = \left| \frac{5k}{5\sqrt{2} \sqrt{25 + k^2}} \right| = \left| \frac{k}{\sqrt{2} \sqrt{25 + k^2}} \right|$.
Squaring both sides,$\frac{1}{4} = \frac{k^2}{2(25 + k^2)}$.
$2(25 + k^2) = 4k^2$.
$50 + 2k^2 = 4k^2$.
$2k^2 = 50 \Rightarrow k^2 = 25$.
Therefore,$k = \pm 5$.

(C) The direction cosines of a line are denoted by $\ell, m, n$.
Given that $\ell = \frac{1}{c}$,$m = \frac{1}{c}$,and $n = \frac{1}{c}$.
We know the fundamental property of direction cosines is $\ell^{2} + m^{2} + n^{2} = 1$.
Substituting the given values: $(\frac{1}{c})^{2} + (\frac{1}{c})^{2} + (\frac{1}{c})^{2} = 1$.
This simplifies to $\frac{1}{c^{2}} + \frac{1}{c^{2}} + \frac{1}{c^{2}} = 1$,which is $\frac{3}{c^{2}} = 1$.
Therefore,$c^{2} = 3$,which gives $c = \pm \sqrt{3}$.

(B) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$.
The sum of the squares of the direction cosines is given by $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Now,we need to evaluate $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1)$
$= 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$
Substituting the value $1$ for the sum of squares:
$= 2(1) - 3 = 2 - 3 = -1$.

(C) The unit vectors along the positive $Y$ and $Z$ axes are $\hat{j} = (0, 1, 0)$ and $\hat{k} = (0, 0, 1)$ respectively.
Any vector along the bisector of the angle between these two axes is given by the sum of these unit vectors: $\vec{v} = \hat{j} + \hat{k} = (0, 1, 1)$.
The magnitude of this vector is $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
The direction cosines are obtained by dividing the components of the vector by its magnitude:
$l = 0/\sqrt{2} = 0$,
$m = 1/\sqrt{2}$,
$n = 1/\sqrt{2}$.
Thus,the direction cosines are $(0, 1/\sqrt{2}, 1/\sqrt{2})$.

(B) We know that the sum of the squares of the direction cosines of a line is $1$,i.e.,$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.
Given that $\alpha + \beta = 90^{\circ}$,we have $\alpha = 90^{\circ} - \beta$.
Substituting this into the expression for $\cos \alpha$,we get $\cos \alpha = \cos(90^{\circ} - \beta) = \sin \beta$.
Therefore,$\cos^{2} \alpha = \sin^{2} \beta$.
Using the identity $\sin^{2} \beta = 1 - \cos^{2} \beta$,we get $\cos^{2} \alpha = 1 - \cos^{2} \beta$,which implies $\cos^{2} \alpha + \cos^{2} \beta = 1$.
Substituting this into the original equation: $(1) + \cos^{2} \gamma = 1$.
This simplifies to $\cos^{2} \gamma = 0$,which means $\cos \gamma = 0$.
Thus,$\gamma = 90^{\circ}$.

(B) Let the direction angles of the line be $\alpha = \frac{\pi}{6}$,$\beta = \frac{\pi}{3}$,and $\gamma$ be the angle with the $Z$ axis.
The sum of the squares of the direction cosines is always $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values: $\cos^2(\frac{\pi}{6}) + \cos^2(\frac{\pi}{3}) + \cos^2 \gamma = 1$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$.
So,$(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1$.
$\frac{3}{4} + \frac{1}{4} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$,which implies $\cos \gamma = 0$.
Therefore,$\gamma = \frac{\pi}{2}$.

(A) Let the angles made by the line with the $x$,$y$,and $z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = 45^{\circ}$ and $\beta = \gamma = \theta$.
The direction cosines are $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 45^{\circ} + \cos^2 \theta + \cos^2 \theta = 1$.
$(\frac{1}{\sqrt{2}})^2 + 2 \cos^2 \theta = 1$.
$\frac{1}{2} + 2 \cos^2 \theta = 1$.
$2 \cos^2 \theta = 1 - \frac{1}{2} = \frac{1}{2}$.
$\cos^2 \theta = \frac{1}{4} \Rightarrow \cos \theta = \pm \frac{1}{2}$.
Thus,the direction cosines are $(\cos 45^{\circ}, \cos \theta, \cos \theta)$,which gives $(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2})$ or $(\frac{1}{\sqrt{2}}, -\frac{1}{2}, -\frac{1}{2})$.

(C) The direction cosines $l, m, n$ of a line must satisfy the condition $l^2 + m^2 + n^2 = 1$.
For option $A$: $(\sqrt{\frac{1}{5}})^2 + (-\sqrt{\frac{1}{2}})^2 + (\sqrt{\frac{3}{10}})^2 = \frac{1}{5} + \frac{1}{2} + \frac{3}{10} = \frac{2+5+3}{10} = \frac{10}{10} = 1$.
For option $B$: $(\frac{1}{\sqrt{3}})^2 + (\frac{-1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$.
For option $C$: $(\frac{1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \neq 1$.
For option $D$: $(\frac{1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 + 0^2 = \frac{1}{2} + \frac{1}{2} + 0 = 1$.
Since the sum of squares for option $C$ is not $1$,it cannot be the direction cosines of a line.

(A) The given equation of the line is $\frac{x+2}{2} = \frac{2y-5}{3}, z = -1$.
First,rewrite the line in standard symmetric form:
$\frac{x+2}{2} = \frac{y - 5/2}{3/2} = \frac{z+1}{0}$.
The direction ratios of the line are $(a, b, c) = (2, 3/2, 0)$.
To simplify,multiply by $2$: $(4, 3, 0)$.
The magnitude of the direction vector is $\sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
The direction cosines $(l, m, n)$ are given by $\left( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \right)$.
Thus,the direction cosines are $\left( \frac{4}{5}, \frac{3}{5}, 0 \right)$.

(B) We know that $\alpha, \beta, \gamma$ are the direction angles of a line,so the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$.
The sum of the squares of the direction cosines is given by $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma = 1$.
We want to evaluate $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$.
Using the identity $\sin ^{2} \theta = 1 - \cos ^{2} \theta$,we have:
$\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma = (1 - \cos ^{2} \alpha) + (1 - \cos ^{2} \beta) + (1 - \cos ^{2} \gamma)$
$= 3 - (\cos ^{2} \alpha + \cos ^{2} \beta + \cos ^{2} \gamma)$
$= 3 - 1 = 2$.

(B) Let the direction cosines of the line $\overrightarrow{OR}$ be $l, m, n$.
Since the line makes angles $\theta_{1}, \theta_{2}, \theta_{3}$ with the planes $XOY, YOZ, ZOX$ respectively,the angles with the normals to these planes (which are the $Z, X, Y$ axes) are $\frac{\pi}{2}-\theta_{1}, \frac{\pi}{2}-\theta_{2}, \frac{\pi}{2}-\theta_{3}$.
Thus,$|l| = \cos(\frac{\pi}{2}-\theta_{2}) = \sin \theta_{2}$,$|m| = \cos(\frac{\pi}{2}-\theta_{3}) = \sin \theta_{3}$,and $|n| = \cos(\frac{\pi}{2}-\theta_{1}) = \sin \theta_{1}$.
We know that for direction cosines,$l^{2}+m^{2}+n^{2}=1$.
Substituting the values,we get $\sin^{2} \theta_{2} + \sin^{2} \theta_{3} + \sin^{2} \theta_{1} = 1$.
Using the identity $\sin^{2} \theta = 1 - \cos^{2} \theta$,we have $(1-\cos^{2} \theta_{2}) + (1-\cos^{2} \theta_{3}) + (1-\cos^{2} \theta_{1}) = 1$.
$3 - (\cos^{2} \theta_{1} + \cos^{2} \theta_{2} + \cos^{2} \theta_{3}) = 1$.
Therefore,$\cos^{2} \theta_{1} + \cos^{2} \theta_{2} + \cos^{2} \theta_{3} = 3 - 1 = 2$.

(C) The direction ratios of the first line are $(a_1, b_1, c_1) = (2, 2, 1)$.
The direction ratios of the line joining $(3, 1, 4)$ and $(7, 2, 12)$ are $(a_2, b_2, c_2) = (7-3, 2-1, 12-4) = (4, 1, 8)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values,we get $\cos \theta = \frac{|(2)(4) + (2)(1) + (1)(8)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}$.
$\cos \theta = \frac{|8 + 2 + 8|}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}} = \frac{18}{\sqrt{9} \sqrt{81}}$.
$\cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{2}{3}\right)$.

(D) The direction ratios of the line $OP$ are $(1, \sqrt{2}, 1)$.
Let the direction cosines of the line $OP$ be $(l, m, n)$.
The magnitude of vector $\vec{OP}$ is $\sqrt{1^2 + (\sqrt{2})^2 + 1^2} = \sqrt{1 + 2 + 1} = \sqrt{4} = 2$.
Thus,the direction cosines are $l = \frac{1}{2}$,$m = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,and $n = \frac{1}{2}$.
The angle $\alpha$ made by a line with direction cosines $(l, m, n)$ with the $YOZ$ plane is given by $\sin \alpha = |l| = \frac{1}{2}$,so $\alpha = 30^{\circ}$.
The angle $\beta$ made with the $ZOX$ plane is given by $\sin \beta = |m| = \frac{1}{\sqrt{2}}$,so $\beta = 45^{\circ}$.
The angle $\gamma$ made with the $XOY$ plane is given by $\sin \gamma = |n| = \frac{1}{2}$,so $\gamma = 30^{\circ}$.
Wait,re-evaluating the question order: $XOY$ plane (involves $z$),$YOZ$ plane (involves $x$),$ZOX$ plane (involves $y$).
Angle with $XOY$ plane: $\sin \theta_1 = |n| = \frac{1}{2} \implies \theta_1 = 30^{\circ}$.
Angle with $YOZ$ plane: $\sin \theta_2 = |l| = \frac{1}{2} \implies \theta_2 = 30^{\circ}$.
Angle with $ZOX$ plane: $\sin \theta_3 = |m| = \frac{1}{\sqrt{2}} \implies \theta_3 = 45^{\circ}$.
Given the options,there might be a convention difference. Let's re-calculate: The angle $\theta$ between a line with direction cosines $(l, m, n)$ and a plane with normal $(a, b, c)$ is $\sin \theta = |al + bm + cn|$.
For $XOY$ ($z=0$,normal $(0,0,1)$): $\sin \theta_1 = |0(1/2) + 0(1/\sqrt{2}) + 1(1/2)| = 1/2 \implies 30^{\circ}$.
For $YOZ$ ($x=0$,normal $(1,0,0)$): $\sin \theta_2 = |1(1/2) + 0 + 0| = 1/2 \implies 30^{\circ}$.
For $ZOX$ ($y=0$,normal $(0,1,0)$): $\sin \theta_3 = |0 + 1(1/\sqrt{2}) + 0| = 1/\sqrt{2} \implies 45^{\circ}$.
Correct answer is $30^{\circ}, 30^{\circ}, 45^{\circ}$.

(C) For a vector $\vec{r}$ to be equally inclined to the coordinate axes, the direction cosines must satisfy $|l| = |m| = |n|$.
Since $l^2 + m^2 + n^2 = 1$, we substitute $|l| = |m| = |n|$ to get $l^2 + l^2 + l^2 = 1$.
This simplifies to $3l^2 = 1$, which gives $l^2 = \frac{1}{3}$.
Thus, $l = \pm \frac{1}{\sqrt{3}}$, $m = \pm \frac{1}{\sqrt{3}}$, and $n = \pm \frac{1}{\sqrt{3}}$.
Each of the direction cosines $l, m, n$ has $2$ possible values $(\pm \frac{1}{\sqrt{3}})$.
Therefore, the total number of such vectors is $2 \times 2 \times 2 = 8$.

(C) Let the vector $v$ make an angle $\alpha$ with each of the three axes. Then the direction cosines of $v$ are $\langle \cos \alpha, \cos \alpha, \cos \alpha \rangle$.
We know that for any vector,the sum of the squares of its direction cosines is $1$:
$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$
$3 \cos^2 \alpha = 1$
$\cos^2 \alpha = \frac{1}{3}$
$\cos \alpha = \pm \frac{1}{\sqrt{3}}$
Therefore,the direction cosines are $\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle$ or $\langle -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \rangle$.

(A) Let the direction angles of the unit vector $\hat{a}$ be $\alpha, \beta, \text{ and } \gamma$.
Given $\alpha = \frac{\pi}{3}$,$\beta = \frac{\pi}{4}$,and $\gamma = \theta$.
We know that for any unit vector,the sum of the squares of the direction cosines is $1$:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
Substituting the given values:
$\cos^2 \left(\frac{\pi}{3}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \theta = 1$
$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1$
$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$
$\frac{3}{4} + \cos^2 \theta = 1$
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$
$\cos \theta = \pm \frac{1}{2}$
Since $\theta \in (0, \pi)$,we have two possible values:
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 given options,$\frac{2\pi}{3}$ is the correct value.

(A) Given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.
First,find the vector $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
$\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.
Next,calculate the magnitude of $\vec{AB}$:
$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
The direction cosines are given by $\frac{x}{|\vec{AB}|}, \frac{y}{|\vec{AB}|}, \frac{z}{|\vec{AB}|}$.
$l = \frac{-2}{6} = -\frac{1}{3}$,$m = \frac{-4}{6} = -\frac{2}{3}$,$n = \frac{4}{6} = \frac{2}{3}$.
Thus,the direction cosines are $(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.

(B) The given equation of the line is $\frac{x-1}{0}=\frac{y+1}{5}=\frac{z-3}{0}$.
Comparing this with the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$,we get the direction ratios as $(a, b, c) = (0, 5, 0)$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$.
Here,$\sqrt{a^2+b^2+c^2} = \sqrt{0^2+5^2+0^2} = \sqrt{25} = 5$.
Thus,$l = \frac{0}{5} = 0$,$m = \frac{5}{5} = 1$,and $n = \frac{0}{5} = 0$.
Therefore,the direction cosines are $(0, 1, 0)$.

(A) The coordinates of the given point are $P(x, y, z) = (6, 7, 8)$.
The perpendicular distance of any point $(x, y, z)$ from the $XY$-plane is given by the absolute value of its $z$-coordinate,which is $|z|$.
Here,the $z$-coordinate is $8$.
Therefore,the perpendicular distance from the $XY$-plane is $|8| = 8$.

(D) The reflection of a point $(x, y, z)$ in the $XY$-plane results in the change of the sign of the $z$-coordinate,while the $x$ and $y$ coordinates remain unchanged.
Therefore,the reflection of the point $(\alpha, \beta, \gamma)$ in the $XY$-plane is $(\alpha, \beta, -\gamma)$.

(A) The signs of the coordinates $(x, y, z)$ in the eight octants are given in the table below:
| Octant | $I$ | $II$ | $III$ | $IV$ | $V$ | $VI$ | $VII$ | $VIII$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $x$ | $+$ | $-$ | $-$ | $+$ | $+$ | $-$ | $-$ | $+$ |
| $y$ | $+$ | $+$ | $-$ | $-$ | $+$ | $+$ | $-$ | $-$ |
| $z$ | $+$ | $+$ | $+$ | $+$ | $-$ | $-$ | $-$ | $-$ |
Given the point $(2, -4, -7)$,we have:
$x = 2$ (positive,$+$)
$y = -4$ (negative,$-$)
$z = -7$ (negative,$-$)
Looking at the table,the octant where $x$ is positive,$y$ is negative,and $z$ is negative is the $VIII$ octant.

(A) The direction cosines of a line are given by $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha, \beta, \gamma$ are the angles made with the $x, y, z$ axes respectively.
Here,$\alpha = 90^{\circ}, \beta = 60^{\circ}, \gamma = \theta$.
The sum of the squares of the direction cosines is always $1$,so $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2 90^{\circ} + \cos^2 60^{\circ} + \cos^2 \theta = 1$.
$(0)^2 + (\frac{1}{2})^2 + \cos^2 \theta = 1$.
$0 + \frac{1}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{1}{4} = \frac{3}{4}$.
Since $\theta$ is acute,$\cos \theta = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$ or $\frac{\pi}{6}$ radians.

(C) Given that the direction angles are $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{3}$.
Let the angle made by the line with the $Z$-axis be $\gamma$.
We know the property of direction cosines: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values:
$\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{3}) + \cos^2 \gamma = 1$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have:
$(\frac{1}{2})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1$.
$\frac{1}{4} + \frac{1}{4} + \cos^2 \gamma = 1$.
$\frac{1}{2} + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 1 - \frac{1}{2} = \frac{1}{2}$.
$\cos \gamma = \pm \frac{1}{\sqrt{2}}$.
Since the angle $\gamma$ is acute,$\cos \gamma = \frac{1}{\sqrt{2}}$,which implies $\gamma = \frac{\pi}{4}$.

(C) Given,direction cosines of line $1$ are $(l_{1}, m_{1}, n_{1}) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$.
Direction cosines of line $2$ are $(l_{2}, m_{2}, n_{2}) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, -\frac{\sqrt{3}}{2}\right)$.
The angle $\theta$ between two lines is given by $\cos \theta = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|$.
Substituting the values:
$\cos \theta = \left|\left(\frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{4}\right) + \left(\frac{1}{4} \times \frac{1}{4}\right) + \left(\frac{\sqrt{3}}{2} \times -\frac{\sqrt{3}}{2}\right)\right|$
$\cos \theta = \left|\frac{3}{16} + \frac{1}{16} - \frac{3}{4}\right|$
$\cos \theta = \left|\frac{3 + 1 - 12}{16}\right| = \left|-\frac{8}{16}\right| = \left|-\frac{1}{2}\right| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) We know that the identity for $\cos 2\theta$ is $\cos 2\theta = 2\cos^2 \theta - 1$.
Substituting this for each term,we get:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1)$.
This simplifies to:
$2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$.
Since $\cos \alpha, \cos \beta, \cos \gamma$ are the direction cosines of a vector $\vec{a}$,we know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting this value into the expression:
$2(1) - 3 = 2 - 3 = -1$.

(A) We know that for a space vector making angles $\alpha, \beta$,and $\gamma$ with the positive direction of the $x, y$,and $z$-axes respectively,the direction cosines satisfy the relation:
$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$
Given that $\alpha = 150^{\circ}$ and $\beta = 60^{\circ}$.
Substituting these values into the equation:
$\cos^{2} 150^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$
Since $\cos 150^{\circ} = -\frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$,we have:
$(-\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} + \cos^{2} \gamma = 1$
$\frac{3}{4} + \frac{1}{4} + \cos^{2} \gamma = 1$
$1 + \cos^{2} \gamma = 1$
$\cos^{2} \gamma = 0$
$\cos \gamma = 0$
Therefore,$\gamma = 90^{\circ}$.

(A) Let the direction cosines of the line be $l, m, n$. Given that the line makes an angle $\alpha = \frac{\pi}{3}$ with both $X$ and $Y$-axes,we have $l = \cos(\frac{\pi}{3}) = \frac{1}{2}$ and $m = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $(\frac{1}{2})^2 + (\frac{1}{2})^2 + n^2 = 1$.
$\frac{1}{4} + \frac{1}{4} + n^2 = 1$ $\Rightarrow \frac{1}{2} + n^2 = 1$ $\Rightarrow n^2 = \frac{1}{2}$.
Thus,$n = \pm \frac{1}{\sqrt{2}}$.
For the acute angle $\gamma$ with the $Z$-axis,$\cos \gamma = |n| = \frac{1}{\sqrt{2}}$.
Therefore,$\gamma = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.

(D) The direction cosines of a ray making angles $\alpha, \beta, \gamma$ with the coordinate axes satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given $\beta = 2\alpha$ and $\gamma = 3\alpha$,we have $\cos^2 \alpha + \cos^2 2\alpha + \cos^2 3\alpha = 1$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,the equation becomes:
$\frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos 4\alpha}{2} + \frac{1 + \cos 6\alpha}{2} = 1$
$3 + \cos 2\alpha + \cos 4\alpha + \cos 6\alpha = 2$
$\cos 2\alpha + \cos 4\alpha + \cos 6\alpha = -1$
$(\cos 6\alpha + \cos 2\alpha) + \cos 4\alpha = -1$
$2 \cos 4\alpha \cos 2\alpha + \cos 4\alpha = -1$
$\cos 4\alpha (2 \cos 2\alpha + 1) = -1$.
Testing the options:
For $\alpha = \frac{\pi}{6}$: $\cos^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} + \cos^2 \frac{\pi}{2} = \frac{3}{4} + \frac{1}{4} + 0 = 1$.
For $\alpha = \frac{\pi}{4}$: $\cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{2} + \cos^2 \frac{3\pi}{4} = \frac{1}{2} + 0 + \frac{1}{2} = 1$.
Thus,the possible values are $\frac{\pi}{6}$ and $\frac{\pi}{4}$.

(C) We know that the sum of the squares of the direction cosines of a vector is $1$.
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
Given that the vector makes equal angles $\alpha$ with $x$ and $y$ axes,so $\alpha = \beta$. Also,it makes $90^{\circ}$ with the $z$-axis,so $\gamma = 90^{\circ}$.
Substituting these values into the identity:
$\cos^2 \alpha + \cos^2 \alpha + \cos^2 90^{\circ} = 1$
$2 \cos^2 \alpha + 0 = 1$
$\cos^2 \alpha = \frac{1}{2}$
$\cos \alpha = \pm \frac{1}{\sqrt{2}}$
Therefore,$\alpha = 45^{\circ}$ or $135^{\circ}$.

(B) Let the point be $P(3, 4, \alpha)$.
The distance of $P$ from the $X$-axis is $d_X = \sqrt{4^2 + \alpha^2} = \sqrt{16 + \alpha^2}$.
The distance of $P$ from the $Y$-axis is $d_Y = \sqrt{3^2 + \alpha^2} = \sqrt{9 + \alpha^2}$.
The distance of $P$ from the $Z$-axis is $d_Z = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
Let $f(\alpha) = \sqrt{16 + \alpha^2} + \sqrt{9 + \alpha^2} + 5$.
To minimize $f(\alpha)$,we find the derivative $f'(\alpha) = \frac{\alpha}{\sqrt{16 + \alpha^2}} + \frac{\alpha}{\sqrt{9 + \alpha^2}}$.
Setting $f'(\alpha) = 0$,we get $\alpha \left( \frac{1}{\sqrt{16 + \alpha^2}} + \frac{1}{\sqrt{9 + \alpha^2}} \right) = 0$.
Since the term in the bracket is always positive,the only solution is $\alpha = 0$.
Thus,$\sec \alpha = \sec(0) = 1$.

(B) Given equations are $l-2m+n=0$ $(1)$ and $lm+10mn-2nl=0$ $(2)$.
From $(1)$,$l = 2m-n$.
Substitute $l$ in $(2)$: $(2m-n)m + 10mn - 2n(2m-n) = 0$.
$2m^2 - mn + 10mn - 4mn + 2n^2 = 0$.
$2m^2 + 5mn + 2n^2 = 0$.
Divide by $n^2$: $2(m/n)^2 + 5(m/n) + 2 = 0$.
$(2m/n + 1)(m/n + 2) = 0$.
Case $1$: $m/n = -1/2 \implies m = -k, n = 2k$.
$l = 2(-k) - 2k = -4k$.
Direction ratios $(l_1, m_1, n_1) = (-4, -1, 2)$.
Case $2$: $m/n = -2 \implies m = -2k, n = k$.
$l = 2(-2k) - k = -5k$.
Direction ratios $(l_2, m_2, n_2) = (-5, -2, 1)$.
$cos \theta = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$.
$cos \theta = \frac{|(-4)(-5) + (-1)(-2) + (2)(1)|}{\sqrt{16+1+4} \sqrt{25+4+1}} = \frac{|20+2+2|}{\sqrt{21} \sqrt{30}} = \frac{24}{\sqrt{630}} = \frac{24}{3\sqrt{70}} = \frac{8}{\sqrt{70}}$.

(A) Let the angles made by the line with the $X, Y$ and $Z$ axes be $\alpha = \frac{\pi}{4}$,$\beta = \frac{\pi}{3}$,and $\gamma = \theta$.
The direction cosines are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $(\cos \frac{\pi}{4})^2 + (\cos \frac{\pi}{3})^2 + \cos^2 \theta = 1$.
$(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + \cos^2 \theta = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1$.
$\frac{3}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $0 < \theta < \frac{\pi}{2}$,$\cos \theta = \frac{1}{2}$.
Thus,the direction cosines are $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos \frac{\pi}{3} = \frac{1}{2}$,and $\cos \theta = \frac{1}{2}$.
Therefore,the direction cosines are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$.

(B) Let the direction ratios of the two lines be $\vec{a} = (2, 2, 1)$ and $\vec{b} = (2, -1, -2)$.
First,we normalize these vectors to find the unit vectors (direction cosines):
$|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$,so $\hat{a} = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$.
$|\vec{b}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3$,so $\hat{b} = (\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3})$.
The angular bisector direction is given by the vector $\vec{v} = \hat{a} + \hat{b}$ or $\vec{v} = \hat{a} - \hat{b}$.
Case $1$: $\vec{v}_1 = (\frac{2}{3} + \frac{2}{3}, \frac{2}{3} - \frac{1}{3}, \frac{1}{3} - \frac{2}{3}) = (\frac{4}{3}, \frac{1}{3}, -\frac{1}{3})$.
Magnitude $|\vec{v}_1| = \sqrt{(\frac{4}{3})^2 + (\frac{1}{3})^2 + (-\frac{1}{3})^2} = \sqrt{\frac{16+1+1}{9}} = \sqrt{2}$.
Direction cosines $(\alpha, \beta, \gamma) = (\frac{4}{3\sqrt{2}}, \frac{1}{3\sqrt{2}}, -\frac{1}{3\sqrt{2}})$.
Then $(\alpha + \beta + \gamma)^2 = (\frac{4+1-1}{3\sqrt{2}})^2 = (\frac{4}{3\sqrt{2}})^2 = \frac{16}{18} = \frac{8}{9}$.
Case $2$: $\vec{v}_2 = (\frac{2}{3} - \frac{2}{3}, \frac{2}{3} + \frac{1}{3}, \frac{1}{3} + \frac{2}{3}) = (0, 1, 1)$.
Magnitude $|\vec{v}_2| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
Direction cosines $(\alpha, \beta, \gamma) = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Then $(\alpha + \beta + \gamma)^2 = (0 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 = (\frac{2}{\sqrt{2}})^2 = (\sqrt{2})^2 = 2$.

(B) Let the line $L_1$ have direction ratios $(1, \alpha, \beta)$,$L_2$ have $(-1, 2, 1)$,and $L_3$ have $(\alpha, 1, \beta)$.
Since $L_1 \perp L_2$,the dot product of their direction ratios is zero:
$1(-1) + \alpha(2) + \beta(1) = 0 \Rightarrow -1 + 2\alpha + \beta = 0 \Rightarrow 2\alpha + \beta = 1$ (Equation $1$).
Since $L_1 \parallel L_3$,their direction ratios are proportional:
$\frac{1}{\alpha} = \frac{\alpha}{1} = \frac{\beta}{\beta}$.
From $\frac{1}{\alpha} = \frac{\alpha}{1}$,we get $\alpha^2 = 1$,so $\alpha = 1$ or $\alpha = -1$.
If $\alpha = 1$,then from Equation $1$: $2(1) + \beta = 1 \Rightarrow \beta = -1$.
If $\alpha = -1$,then from Equation $1$: $2(-1) + \beta = 1 \Rightarrow \beta = 3$.
However,the condition $\frac{\beta}{\beta} = 1$ must hold for $\beta \neq 0$. If $\beta = 0$,the ratio is undefined. Checking $\alpha = 1, \beta = -1$: the ratios are $(1, 1, -1)$ and $(1, 1, -1)$,which are parallel. Thus,$(\alpha, \beta) = (1, -1)$.

(D) Let the direction cosines of line $L$ be $(l, m, n)$.
Given that $L$ makes angles $\pi / 3$ and $\pi / 4$ with $Y$-axis and $Z$-axis respectively.
Therefore,$m = \cos(\pi / 3) = 1 / 2$ and $n = \cos(\pi / 4) = 1 / \sqrt{2}$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,$l^2 + (1 / 2)^2 + (1 / \sqrt{2})^2 = 1 \Rightarrow l^2 + 1 / 4 + 1 / 2 = 1 \Rightarrow l^2 = 1 - 3 / 4 = 1 / 4$.
Thus,$l = 1 / 2$ (taking the positive value).
The direction ratios of the second line are $(1, 1, 1)$. Its direction cosines are $(1 / \sqrt{3}, 1 / \sqrt{3}, 1 / \sqrt{3})$.
The angle $\theta$ between the two lines is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(1 / 2)(1 / \sqrt{3}) + (1 / 2)(1 / \sqrt{3}) + (1 / \sqrt{2})(1 / \sqrt{3})| = |1 / (2 \sqrt{3}) + 1 / (2 \sqrt{3}) + 1 / \sqrt{6}|$.
$\cos \theta = |1 / \sqrt{3} + 1 / \sqrt{6}| = |\sqrt{2} / \sqrt{6} + 1 / \sqrt{6}| = (\sqrt{2} + 1) / \sqrt{6}$.
Therefore,$\theta = \cos^{-1} \left( \frac{\sqrt{2} + 1}{\sqrt{6}} \right)$.

(B) Given that $(l, m, n)$ are the direction cosines of a line perpendicular to two lines with direction ratios $(1, 2, -1)$ and $(1, -2, 1)$.
Since the line is perpendicular to both,we have:
$l + 2m - n = 0$ ...$(i)$
$l - 2m + n = 0$ ...$(ii)$
Adding $(i)$ and $(ii)$,we get $2l = 0$,which implies $l = 0$.
Substituting $l = 0$ in $(i)$,we get $2m - n = 0$,so $n = 2m$.
We know that for direction cosines,$l^2 + m^2 + n^2 = 1$.
Substituting $l = 0$ and $n = 2m$,we get $0^2 + m^2 + (2m)^2 = 1$.
$m^2 + 4m^2 = 1 \Rightarrow 5m^2 = 1 \Rightarrow m^2 = \frac{1}{5}$.
Now,we need to find $(l + m + n)^2$.
$(l + m + n)^2 = (0 + m + 2m)^2 = (3m)^2 = 9m^2$.
Substituting $m^2 = \frac{1}{5}$,we get $(l + m + n)^2 = 9 \times \frac{1}{5} = \frac{9}{5}$.

(B) The direction ratios of $AB$ are $a_1 = 6-1 = 5$,$b_1 = 11-(-1) = 12$,$c_1 = 2-2 = 0$.
The magnitude of $AB$ is $\sqrt{5^2+12^2+0^2} = \sqrt{25+144} = \sqrt{169} = 13$.
Thus,the direction cosines of $AB$ are $l_1 = \frac{5}{13}$,$m_1 = \frac{12}{13}$,$n_1 = 0$.
The direction ratios of $AC$ are $a_2 = 1-1 = 0$,$b_2 = 2-(-1) = 3$,$c_2 = 6-2 = 4$.
The magnitude of $AC$ is $\sqrt{0^2+3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
Thus,the direction cosines of $AC$ are $l_2 = 0$,$m_2 = \frac{3}{5}$,$n_2 = \frac{4}{5}$.
The value of $|l_1 l_2 + m_1 m_2 + n_1 n_2|$ is the cosine of the angle between the lines $AB$ and $AC$.
$|l_1 l_2 + m_1 m_2 + n_1 n_2| = |(\frac{5}{13} \times 0) + (\frac{12}{13} \times \frac{3}{5}) + (0 \times \frac{4}{5})| = |0 + \frac{36}{65} + 0| = \frac{36}{65}$.

(B) Given that $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines.
Since they are direction cosines,we have $l_1^2 + m_1^2 + n_1^2 = 1$ and $l_2^2 + m_2^2 + n_2^2 = 1$.
Let $\theta$ be the angle between the two lines. Then $\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$.
The expression is $(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 l_2 + m_1 m_2 + n_1 n_2)^2$.
By Lagrange's Identity,$(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 = (l_1^2 + m_1^2 + n_1^2)(l_2^2 + m_2^2 + n_2^2) - (l_1 l_2 + m_1 m_2 + n_1 n_2)^2$.
Substituting the values,we get $(1)(1) - \cos^2 \theta = \sin^2 \theta$.
Thus,the total expression becomes $\sin^2 \theta + \cos^2 \theta = 1$.

(B) Let the angles made by the line $L$ with the positive $X$,$Y$,and $Z$ axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given that $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$.
The relationship between direction cosines is given by $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values:
$\cos^2 \frac{\pi}{3} + \cos^2 \frac{\pi}{4} + \cos^2 \gamma = 1$
$\Rightarrow (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$
$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$
$\Rightarrow \frac{3}{4} + \cos^2 \gamma = 1$
$\Rightarrow \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$
$\Rightarrow \cos \gamma = \pm \frac{1}{2}$.
Since the question asks for the angle with the positive direction of the $Z$-axis,we take $\cos \gamma = \frac{1}{2}$.
Therefore,$\gamma = \frac{\pi}{3}$.