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

(B) The given equation of the line is $6x - 2 = 3y + 1 = 2z - 2$.
Rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$:
$6(x - \frac{1}{3}) = 3(y + \frac{1}{3}) = 2(z - 1)$
Dividing by the least common multiple of $6, 3, 2$,which is $6$:
$\frac{x - 1/3}{1/6} = \frac{y + 1/3}{1/3} = \frac{z - 1}{1/2}$
Multiplying the denominators by $6$ to get direction ratios $(a, b, c)$:
$a = 1, b = 2, c = 3$
The magnitude of the direction vector is $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}})$.
Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.

(C) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.
Given $\alpha = 45^\circ$ and $\beta = 120^\circ$.
The direction cosines are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $\cos^2(45^\circ) + \cos^2(120^\circ) + \cos^2(\gamma) = 1$.
$(\frac{1}{\sqrt{2}})^2 + (-\frac{1}{2})^2 + \cos^2(\gamma) = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^2(\gamma) = 1$.
$\frac{3}{4} + \cos^2(\gamma) = 1$.
$\cos^2(\gamma) = 1 - \frac{3}{4} = \frac{1}{4}$.
$\cos(\gamma) = \pm \frac{1}{2}$.
Since $\theta$ is an acute angle,$\cos(\theta) = \frac{1}{2}$.
Therefore,$\theta = 60^\circ$.

(A) Let the projections of the line segment on the $x, y, z$ axes be $a = 12$,$b = 4$,and $c = 3$.
The length of the line segment $L$ is given by the formula $L = \sqrt{a^2 + b^2 + c^2}$.
Substituting the values: $L = \sqrt{12^2 + 4^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.
The direction cosines $(l, m, n)$ are given by $l = a/L$,$m = b/L$,and $n = c/L$.
Therefore,$l = 12/13$,$m = 4/13$,and $n = 3/13$.
Thus,the length is $13$ and the direction cosines are $< 12/13, 4/13, 3/13 >$.

(B) The projection of a line segment $AB$ on another line $CD$ is given by the length of the segment $AB$ multiplied by the cosine of the angle $\theta$ between them.
Thus,the projection = $AB \cos \theta$.

(D) The direction ratios of the line passing through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ are proportional to $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
For points $A(6, -7, -1)$ and $B(2, -3, 1)$,the direction ratios are $(2 - 6, -3 - (-7), 1 - (-1)) = (-4, 4, 2)$.
Dividing by $2$,we get the simplified direction ratios as $(-2, 2, 1)$.
The magnitude of the vector is $\sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The direction cosines are $\pm \frac{-2}{3}, \pm \frac{2}{3}, \pm \frac{1}{3}$.
The angle $\alpha$ made with the $x$-axis is acute,which means $\cos \alpha > 0$. The direction cosine corresponding to the $x$-axis is $l = \pm \frac{-2}{3}$.
To satisfy $\cos \alpha > 0$,we must choose the negative sign for the direction ratios so that $l = -(\frac{-2}{3}) = \frac{2}{3}$.
Thus,the direction cosines are $(\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3})$.

(B) Let the points be $P(4, 3, -5)$ and $Q(-2, 1, -8)$.
First,we find the direction ratios $(a, b, c)$ of the line $PQ$:
$a = x_2 - x_1 = -2 - 4 = -6$
$b = y_2 - y_1 = 1 - 3 = -2$
$c = z_2 - z_1 = -8 - (-5) = -3$
Alternatively,taking the vector $QP$:
$a = 4 - (-2) = 6, b = 3 - 1 = 2, c = -5 - (-8) = 3$.
The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$.
Thus,$(l, m, n) = (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.

(A) In a $3D$ Cartesian coordinate system,the projection of a point $(x, y, z)$ onto a coordinate plane is obtained by setting the coordinate corresponding to the axis perpendicular to that plane to $0$.
For the $yz$-plane,the $x$-coordinate is perpendicular to the plane.
Therefore,the projection of the point $(a, b, c)$ onto the $yz$-plane is obtained by setting the $x$-coordinate to $0$.
Thus,the projection is $(0, b, c)$.

(D) The direction ratios of the line are $a = 6, b = 2, c = 3$.
The direction cosines $(l, m, n)$ are given by:
$l = \frac{6}{\sqrt{6^2 + 2^2 + 3^2}} = \frac{6}{\sqrt{36 + 4 + 9}} = \frac{6}{\sqrt{49}} = \frac{6}{7}$
$m = \frac{2}{\sqrt{49}} = \frac{2}{7}$
$n = \frac{3}{\sqrt{49}} = \frac{3}{7}$
The projection of the line segment joining points $P(x_1, y_1, z_1) = (-1, 0, 3)$ and $Q(x_2, y_2, z_2) = (2, 5, 1)$ on the line is given by:
$Projection = l(x_2 - x_1) + m(y_2 - y_1) + n(z_2 - z_1)$
Substituting the values:
$Projection = \frac{6}{7}(2 - (-1)) + \frac{2}{7}(5 - 0) + \frac{3}{7}(1 - 3)$
$Projection = \frac{6}{7}(3) + \frac{2}{7}(5) + \frac{3}{7}(-2)$
$Projection = \frac{18}{7} + \frac{10}{7} - \frac{6}{7}$
$Projection = \frac{18 + 10 - 6}{7} = \frac{22}{7}$

(A) Given direction ratios are $a = 1, b = -3, c = 2$.
The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.
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 $\left( \frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right)$.

(B) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$.
It is a known property that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Now,we use the identity $\cos 2\theta = 2 \cos^2 \theta - 1$.
Substituting this into the expression:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2 \cos^2 \alpha - 1) + (2 \cos^2 \beta - 1) + (2 \cos^2 \gamma - 1)$
$= 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$
$= 2(1) - 3$
$= 2 - 3 = -1$.

(A) Let the direction cosines of the vector $\overline{OP}$ be $l, m, n$.
Given that the angles with the axes are $\alpha = 45^{\circ}$,$\beta = 60^{\circ}$,and let the angle with $OZ$ be $\gamma$.
Then,$l = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,$m = \cos(60^{\circ}) = \frac{1}{2}$,and $n = \cos(\gamma)$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + n^2 = 1$.
$\frac{1}{2} + \frac{1}{4} + n^2 = 1$.
$\frac{3}{4} + n^2 = 1 \Rightarrow n^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
Thus,$n = \pm \frac{1}{2}$.
Since $n = \cos(\gamma)$,we have $\cos(\gamma) = \frac{1}{2}$ or $\cos(\gamma) = -\frac{1}{2}$.
Therefore,$\gamma = 60^{\circ}$ or $\gamma = 120^{\circ}$.

(B) Let the direction cosines of the required line be $(l, m, n)$. Since the line makes equal angles $\theta$ with the three mutually perpendicular lines,we have:
$l \cdot l_1 + m \cdot m_1 + n \cdot n_1 = \cos \theta$
$l \cdot l_2 + m \cdot m_2 + n \cdot n_2 = \cos \theta$
$l \cdot l_3 + m \cdot m_3 + n \cdot n_3 = \cos \theta$
Squaring and adding these equations,we get:
$(l \cdot l_1 + m \cdot m_1 + n \cdot n_1)^2 + (l \cdot l_2 + m \cdot m_2 + n \cdot n_2)^2 + (l \cdot l_3 + m \cdot m_3 + n \cdot n_3)^2 = 3 \cos^2 \theta$
Using the property of direction cosines for mutually perpendicular lines,this simplifies to $l^2 + m^2 + n^2 = 3 \cos^2 \theta$. Since $l^2 + m^2 + n^2 = 1$,we have $3 \cos^2 \theta = 1$,so $\cos \theta = \frac{1}{\sqrt{3}}$.
Thus,$l = \frac{l_1 + l_2 + l_3}{\sqrt{3}}$,$m = \frac{m_1 + m_2 + m_3}{\sqrt{3}}$,$n = \frac{n_1 + n_2 + n_3}{\sqrt{3}}$.

(C) Let the points be $A(-1, 2, 3)$ and $B(-1, 4, 0)$.
The direction ratios of the line segment $AB$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-1 - (-1), 4 - 2, 0 - 3) = (0, 2, -3)$.
The line makes angles $\alpha = 45^{\circ}, \beta = 60^{\circ}, \gamma = 60^{\circ}$ with the coordinate axes.
The direction cosines of the line are $l = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,$m = \cos(60^{\circ}) = \frac{1}{2}$,and $n = \cos(60^{\circ}) = \frac{1}{2}$.
The projection of a line segment with direction ratios $(a, b, c)$ on a line with direction cosines $(l, m, n)$ is given by $|al + bm + cn|$.
Substituting the values: Projection $= |(0)(\frac{1}{\sqrt{2}}) + (2)(\frac{1}{2}) + (-3)(\frac{1}{2})| = |0 + 1 - \frac{3}{2}| = |-\frac{1}{2}| = \frac{1}{2}$.

(C) Let the points be $A(5, 2, 3)$ and $B(-1, 0, 2)$.
The vector $\vec{AB}$ is given by $(-1-5)\hat{i} + (0-2)\hat{j} + (2-3)\hat{k} = -6\hat{i} - 2\hat{j} - \hat{k}$.
The direction ratios of the line are $1, 2, 3$,so the direction vector of the line is $\vec{v} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.
The unit vector along the line is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{14}}$.
The projection of the line segment $AB$ on the line is the absolute value of the dot product of $\vec{AB}$ and $\hat{u}$.
Projection $= |\vec{AB} \cdot \hat{u}| = |(-6\hat{i} - 2\hat{j} - \hat{k}) \cdot \frac{\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{14}}|$
$= |\frac{(-6)(1) + (-2)(2) + (-1)(3)}{\sqrt{14}}| = |\frac{-6 - 4 - 3}{\sqrt{14}}| = |\frac{-13}{\sqrt{14}}| = \frac{13}{\sqrt{14}}$.

(B) In a three-dimensional Cartesian coordinate system,any point $P$ is represented by the coordinates $(x, y, z)$.
For a point to lie on the $z$-axis,its distance from both the $x$-axis and the $y$-axis must be zero.
This implies that the $x$-coordinate must be $0$ and the $y$-coordinate must be $0$.
Therefore,any point on the $z$-axis is of the form $(0, 0, z)$,where $z$ is any real number.
Thus,the condition satisfied is that both $x$ and $y$ coordinates are zero.

(A) Given the relations:
$l + m + n = 0 \implies n = -(l + m)$
Substitute $n$ into the second equation:
$l^2 + m^2 - (-(l + m))^2 = 0$
$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$
$-2lm = 0 \implies lm = 0$
This implies either $l = 0$ or $m = 0$.
Case $1$: If $l = 0$,then $m + n = 0 \implies n = -m$. The direction ratios are $(0, m, -m)$,which simplifies to $(0, 1, -1)$.
Case $2$: If $m = 0$,then $l + n = 0 \implies n = -l$. The direction ratios are $(l, 0, -l)$,which simplifies to $(1, 0, -1)$.
Let the angle between the lines be $\theta$. The direction ratios are $\vec{a} = (0, 1, -1)$ and $\vec{b} = (1, 0, -1)$.
$\cos \theta = \frac{|(0)(1) + (1)(0) + (-1)(-1)|}{\sqrt{0^2 + 1^2 + (-1)^2} \sqrt{1^2 + 0^2 + (-1)^2}}$
$\cos \theta = \frac{|0 + 0 + 1|}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
Since the angle between lines is usually taken as the acute angle,$\cos \theta = 1/2 \implies \theta = \pi / 3$.
However,if we consider the direction cosines as vectors,the angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{-1}{\sqrt{2}\sqrt{2}} = -1/2$,which gives $\theta = 2\pi / 3$.

(B) The direction cosines of a line are $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we can write:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2 \cos^2 \alpha - 1) + (2 \cos^2 \beta - 1) + (2 \cos^2 \gamma - 1)$
$= 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$
$= 2(1) - 3$
$= 2 - 3 = -1$.

(A) The projections of a line on the coordinate axes are the direction ratios of the line,denoted as $a = 4$,$b = 6$,and $c = 12$.
To find the direction cosines $(l, m, n)$,we use the formula $l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$,$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$,and $n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.
First,calculate the magnitude: $\sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.
Now,calculate the direction cosines:
$l = \frac{4}{14} = \frac{2}{7}$
$m = \frac{6}{14} = \frac{3}{7}$
$n = \frac{12}{14} = \frac{6}{7}$
Thus,the direction cosines are $(\frac{2}{7}, \frac{3}{7}, \frac{6}{7})$.

(A) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$.
Therefore,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we can write:
$(1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1$.
$3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1$.
$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2$.

(B) Let the line make an angle $\alpha$ with each of the coordinate axes.
Thus,the direction cosines are $l = \cos \alpha$,$m = \cos \alpha$,and $n = \cos \alpha$.
We know that for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1 \implies \cos^2 \alpha = \frac{1}{3}$.
Therefore,$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction cosines are $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ or $\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$.

(D) Let the direction cosines of the line be $(l, m, n)$.
Given that the line makes an angle of $\alpha = \pi /4$ with the $x$-axis and $\beta = \pi /4$ with the $y$-axis.
Thus,$l = \cos(\pi /4) = 1/\sqrt{2}$ and $m = \cos(\pi /4) = 1/\sqrt{2}$.
We know the identity for direction cosines: $l^2 + m^2 + n^2 = 1$.
Substituting the values: $(1/\sqrt{2})^2 + (1/\sqrt{2})^2 + n^2 = 1$.
$1/2 + 1/2 + n^2 = 1$.
$1 + n^2 = 1 \implies n^2 = 0 \implies n = 0$.
If $n = \cos(\gamma) = 0$,then $\gamma = \pi /2$.
Therefore,the line makes an angle of $\pi /2$ with the positive direction of the $z$-axis.

(B) Let the angles made by the line with the $x, y,$ and $z$ axes be $\alpha, \beta,$ and $\gamma$ respectively.
Given $\alpha = 120^{\circ}$ and $\beta = 60^{\circ}$.
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:
$\cos^{2} (120^{\circ}) + \cos^{2} (60^{\circ}) + \cos^{2} \gamma = 1$
Since $\cos(120^{\circ}) = -\frac{1}{2}$ and $\cos(60^{\circ}) = \frac{1}{2}$,we have:
$(-\frac{1}{2})^{2} + (\frac{1}{2})^{2} + \cos^{2} \gamma = 1$
$\frac{1}{4} + \frac{1}{4} + \cos^{2} \gamma = 1$
$\frac{1}{2} + \cos^{2} \gamma = 1$
$\cos^{2} \gamma = 1 - \frac{1}{2} = \frac{1}{2}$
$\cos \gamma = \pm \frac{1}{\sqrt{2}}$
Therefore,$\gamma = 45^{\circ}$ or $135^{\circ}$.

(C) Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel if and only if their direction ratios are proportional.
Mathematically,this condition is expressed as:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k$
where $k$ is a non-zero constant.
This implies that the direction vectors of the two lines,$\vec{v_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{v_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$,are collinear,meaning $\vec{v_1} = k\vec{v_2}$.
Therefore,the correct condition for the lines to be parallel is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

(A) Let the direction ratios of the three lines be $L_1 = (1, 1, 2)$,$L_2 = (\sqrt{3}-1, -\sqrt{3}-1, 4)$,and $L_3 = (-\sqrt{3}-1, \sqrt{3}-1, 4)$.
To find the angles between these lines,we use the formula $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.
First,calculate the magnitudes: $|L_1| = \sqrt{1^2+1^2+2^2} = \sqrt{6}$.
$|L_2| = \sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3+1-2\sqrt{3}) + (3+1+2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$.
$|L_3| = \sqrt{(-\sqrt{3}-1)^2 + (\sqrt{3}-1)^2 + 4^2} = \sqrt{(3+1+2\sqrt{3}) + (3+1-2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$.
Now,calculate the dot products:
$L_1 \cdot L_2 = 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3}-1-\sqrt{3}-1+8 = 6$.
$L_1 \cdot L_3 = 1(-\sqrt{3}-1) + 1(\sqrt{3}-1) + 2(4) = -\sqrt{3}-1+\sqrt{3}-1+8 = 6$.
$L_2 \cdot L_3 = (\sqrt{3}-1)(-\sqrt{3}-1) + (-\sqrt{3}-1)(\sqrt{3}-1) + 4(4) = -(3-1) - (3-1) + 16 = -2 - 2 + 16 = 12$.
Calculate the cosines of the angles:
$\cos \theta_{12} = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \implies \theta_{12} = 60^\circ$.
$\cos \theta_{13} = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \implies \theta_{13} = 60^\circ$.
$\cos \theta_{23} = \frac{12}{2\sqrt{6} \cdot 2\sqrt{6}} = \frac{12}{24} = \frac{1}{2} \implies \theta_{23} = 60^\circ$.
Since all angles are $60^\circ$,the lines form an equilateral triangle.

(C) The projection of a line segment joining points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ on a line with direction cosines $(l, m, n)$ is given by the formula:
$|l(x_2 - x_1) + m(y_2 - y_1) + n(z_2 - z_1)|$.
Given points are $P(7, -5, 11)$ and $Q(-2, 8, 13)$.
Direction cosines are $l = \frac{1}{3}, m = \frac{2}{3}, n = \frac{2}{3}$.
Calculating the differences:
$x_2 - x_1 = -2 - 7 = -9$
$y_2 - y_1 = 8 - (-5) = 13$
$z_2 - z_1 = 13 - 11 = 2$
Substituting these values into the formula:
Projection $= |(\frac{1}{3})(-9) + (\frac{2}{3})(13) + (\frac{2}{3})(2)|$
$= |-\frac{9}{3} + \frac{26}{3} + \frac{4}{3}|$
$= |\frac{-9 + 26 + 4}{3}|$
$= |\frac{21}{3}| = 7$.

(A) Let the coordinates of the point be $P(x, y, z)$.
The distance of point $P$ from the $xy$-plane is $|z|$.
The distance of point $P$ from the $yz$-plane is $|x|$.
The distance of point $P$ from the $zx$-plane is $|y|$.
According to the problem,the sum of the distances from the $xy$-plane and the $yz$-plane is equal to the distance from the $zx$-plane:
$|z| + |x| = |y|$.
Assuming the point lies in the first octant where $x, y, z > 0$,we have $x + z = y$,which simplifies to $x - y + z = 0$.

(C) Let the direction ratios of the two lines be $\vec{a} = (1, 1, 2)$ and $\vec{b} = (\sqrt{3} - 1, -\sqrt{3} - 1, 4)$.
The formula for 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 $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
First,calculate the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 1(\sqrt{3} - 1) + 1(-\sqrt{3} - 1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.
Next,calculate the magnitudes:
$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
$|\vec{b}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
Now,substitute these values into the formula:
$\cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{2 \times 6} = \frac{6}{12} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^o$.

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$.
Given: $(a_1, b_1, c_1) = (1, 1, 2)$ and $(a_2, b_2, c_2) = (\sqrt{3}-1, -\sqrt{3}-1, 4)$.
The formula for the angle $\theta$ between two lines is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
First,calculate the dot product:
$a_1 a_2 + b_1 b_2 + c_1 c_2 = 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.
Next,calculate the magnitudes:
$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}$.
$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
Now,substitute these values into the formula:
$\cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{2 \times 6} = \frac{6}{12} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ}$.

(A) Given direction ratios are $a = 1, b = -3, c = 2$.
The formula for direction cosines $(l, m, n)$ is given by:
$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.
First,calculate the magnitude:
$\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.
Now,calculate the direction cosines:
$l = \frac{1}{\sqrt{14}}$,
$m = \frac{-3}{\sqrt{14}}$,
$n = \frac{2}{\sqrt{14}}$.
Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}}$.

(A) Let the direction cosines of the line be $l, m, n$. Since the line makes an angle $\theta$ with $X$ and $Z$ axes,we have $l = \cos \theta$ and $n = \cos \theta$.
Since the line makes an angle $\beta$ with the $Y$ axis,we have $m = \cos \beta$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \theta + \cos^2 \beta + \cos^2 \theta = 1$.
$2 \cos^2 \theta + \cos^2 \beta = 1$.
Using the identity $\sin^2 \beta = 1 - \cos^2 \beta$,we have $\cos^2 \beta = 1 - \sin^2 \beta$.
Substituting this into the equation: $2 \cos^2 \theta + (1 - \sin^2 \beta) = 1$.
$2 \cos^2 \theta = \sin^2 \beta$.
Given $\sin^2 \beta = 3 \sin^2 \theta$,we substitute this into the equation:
$2 \cos^2 \theta = 3 \sin^2 \theta$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$:
$2 \cos^2 \theta = 3(1 - \cos^2 \theta)$.
$2 \cos^2 \theta = 3 - 3 \cos^2 \theta$.
$5 \cos^2 \theta = 3$.
$\cos^2 \theta = 3/5$.

(A) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$.
Here,$(a_1, b_1, c_1) = \left( \frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2} \right)$ and $(a_2, b_2, c_2) = \left( \frac{\sqrt{3}}{4}, \frac{1}{4}, -\frac{\sqrt{3}}{2} \right)$.
First,we calculate the magnitudes of the vectors:
$|v_1| = \sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{3}{16} + \frac{1}{16} + \frac{3}{4}} = \sqrt{\frac{3+1+12}{16}} = \sqrt{\frac{16}{16}} = 1$.
$|v_2| = \sqrt{(\frac{\sqrt{3}}{4})^2 + (\frac{1}{4})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{3}{16} + \frac{1}{16} + \frac{3}{4}} = 1$.
Since the magnitudes are $1$,the given ratios are actually direction cosines.
The cosine of the angle $\theta$ between the lines is given by $\cos \theta = |a_1a_2 + b_1b_2 + c_1c_2|$.
$\cos \theta = |(\frac{\sqrt{3}}{4})(\frac{\sqrt{3}}{4}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2})|$
$\cos \theta = |\frac{3}{16} + \frac{1}{16} - \frac{3}{4}| = |\frac{4}{16} - \frac{12}{16}| = |-\frac{8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.
Thus,$\theta = \cos^{-1}(\frac{1}{2}) = 60^o$.
However,if we consider the angle between the vectors,$\cos \theta = -\frac{1}{2}$,which gives $\theta = 120^o$.

(A) The equation of the $yz$-plane is $x = 0$.
Any line perpendicular to the $yz$-plane must be parallel to the $x$-axis.
The direction ratios of the $x$-axis are $(1, 0, 0)$.
Since the line is parallel to the $x$-axis,its direction ratios are proportional to $(1, 0, 0)$.
Therefore,the direction cosines of the line are $(1, 0, 0)$.

(D) 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 for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1$.
$\cos^2 \alpha = 1/3$.
$\cos \alpha = 1/\sqrt{3}$ (taking the positive value for the acute angle).
Therefore,$\alpha = \cos^{-1}(1/\sqrt{3})$.

(C) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know the fundamental identity for direction cosines is $l^2 + m^2 + n^2 = 1$.
Given $l = \cos \alpha = 14/15$ and $m = \cos \beta = 1/3$.
Substituting these values into the identity:
$(14/15)^2 + (1/3)^2 + \cos^2 \gamma = 1$
$196/225 + 1/9 + \cos^2 \gamma = 1$
To add the fractions,find a common denominator,which is $225$:
$196/225 + 25/225 + \cos^2 \gamma = 1$
$221/225 + \cos^2 \gamma = 1$
$\cos^2 \gamma = 1 - 221/225$
$\cos^2 \gamma = (225 - 221) / 225$
$\cos^2 \gamma = 4/225$
Taking the square root on both sides:
$\cos \gamma = \pm \sqrt{4/225} = \pm 2/15$.

(D) The coordinates of point $P$ are $(x, y, z) = (3, 12, 4)$.
The distance $OP = \sqrt{x^2 + y^2 + z^2} = \sqrt{3^2 + 12^2 + 4^2}$.
$OP = \sqrt{9 + 144 + 16} = \sqrt{169} = 13$.
The direction cosines $(l, m, n)$ are given by $\frac{x}{OP}, \frac{y}{OP}, \frac{z}{OP}$.
Therefore,the direction cosines are $\frac{3}{13}, \frac{12}{13}, \frac{4}{13}$.

(B) Let the two lines be $L_1$ and $L_2$ with direction ratios $\langle a, b, c \rangle$ and $\langle a', b', c' \rangle$ respectively.
Two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ are perpendicular if and only if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Since the given lines are perpendicular,the condition $aa' + bb' + cc' = 0$ must hold true.
Therefore,the correct option is $B$.

(A) Let the vector be $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$.
Given that the projections of the vector on the coordinate axes are $6, -3, 2$,we have $a = 6$,$b = -3$,and $c = 2$.
The magnitude of the vector is $|\vec{r}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{6^2 + (-3)^2 + 2^2}$.
$|\vec{r}| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
The direction cosines $(l, m, n)$ of a vector are given by $\frac{a}{|\vec{r}|}, \frac{b}{|\vec{r}|}, \frac{c}{|\vec{r}|}$.
Thus,$l = \frac{6}{7}$,$m = \frac{-3}{7}$,and $n = \frac{2}{7}$.
Therefore,the direction cosines are $\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}$.

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1 = (3, 4, 5)$ and $a_2, b_2, c_2 = (4, -3, 5)$.
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 the formula:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the values:
$a_1 a_2 + b_1 b_2 + c_1 c_2 = (3)(4) + (4)(-3) + (5)(5) = 12 - 12 + 25 = 25$
$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$
$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + (-3)^2 + 5^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$
Therefore,$\cos \theta = \frac{25}{(5\sqrt{2})(5\sqrt{2})} = \frac{25}{25 \times 2} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^o$.

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1 = 1, 1, 2$ and $a_2, b_2, c_2 = \sqrt{3}-1, -\sqrt{3}-1, 4$.
The cosine of the angle $\theta$ between the lines is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Calculating the numerator:
$1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.
Calculating the denominators:
$\sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$.
$\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
Thus,$\cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{2 \cdot 6} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

(A) The direction ratios of the line passing through $A(6, -7, -1)$ and $B(2, -3, 1)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 6, -3 - (-7), 1 - (-1)) = (-4, 4, 2)$.
The magnitude of the vector is $\sqrt{(-4)^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.
The direction cosines $(l, m, n)$ are $\left( \frac{-4}{6}, \frac{4}{6}, \frac{2}{6} \right) = \left( -\frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right)$ or $\left( \frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right)$.
Let $\alpha$ be the angle made by the line with the positive $x$-axis. Then $\cos \alpha = l$. For $\alpha$ to be an acute angle,$\cos \alpha > 0$,which implies $l > 0$.
Comparing the two sets of direction cosines,the set with $l > 0$ is $\left( \frac{2}{3}, -\frac{2}{3}, -\frac{1}{3} \right)$.

(B) For a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines satisfy the relation: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given that $\alpha + \beta = 90^\circ$,we have $\beta = 90^\circ - \alpha$.
Substituting this into the relation: $\cos^2 \alpha + \cos^2(90^\circ - \alpha) + \cos^2 \gamma = 1$.
Since $\cos(90^\circ - \alpha) = \sin \alpha$,the equation becomes: $\cos^2 \alpha + \sin^2 \alpha + \cos^2 \gamma = 1$.
Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get: $1 + \cos^2 \gamma = 1$.
This implies $\cos^2 \gamma = 0$,so $\cos \gamma = 0$.
Therefore,$\gamma = 90^\circ$.

(A) Let the length of the line segment be $r$ and its direction cosines be $l, m, n$.
The projections of the line segment on the coordinate axes are given by $lr, mr, nr$.
Given that $lr = 2, mr = 3$,and $nr = 6$.
Squaring and adding these equations,we get:
$(lr)^2 + (mr)^2 + (nr)^2 = 2^2 + 3^2 + 6^2$
$r^2(l^2 + m^2 + n^2) = 4 + 9 + 36$
Since the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$,we have:
$r^2(1) = 49$
$r^2 = 49$
$r = 7$
Thus,the length of the line segment is $7$.

(B) Let the direction ratios of the two lines be $(a_1, b_1, c_1) = (2, 3, 6)$ and $(a_2, b_2, c_2) = (1, 2, 2)$.
The formula for 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:
$\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$
Substituting the values:
$\cos \theta = \left| \frac{(2)(1) + (3)(2) + (6)(2)}{\sqrt{2^2 + 3^2 + 6^2} \sqrt{1^2 + 2^2 + 2^2}} \right|$
Calculating the numerator:
$2 + 6 + 12 = 20$
Calculating the denominators:
$\sqrt{4 + 9 + 36} = \sqrt{49} = 7$
$\sqrt{1 + 4 + 4} = \sqrt{9} = 3$
Thus,$\cos \theta = \left| \frac{20}{7 \times 3} \right| = \frac{20}{21}$.
Therefore,$\theta = \cos ^{ - 1} \left( \frac{20}{21} \right)$.

(B) The projection of a line segment joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ on a line with direction cosines $l, m, n$ is given by the formula:
Projection = $l(x_2 - x_1) + m(y_2 - y_1) + n(z_2 - z_1)$.
This is derived from the dot product of the vector $\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$ and the unit vector along the line $AB$,which is $\hat{u} = l\hat{i} + m\hat{j} + n\hat{k}$.
Thus,the projection is $\vec{PQ} \cdot \hat{u} = l(x_2 - x_1) + m(y_2 - y_1) + n(z_2 - z_1)$.

(A) Let the direction cosines of the vector $\vec{r}$ be $l, m, n$. The projections of the vector on the coordinate axes are given by $l|\vec{r}|, m|\vec{r}|, n|\vec{r}|$.
Therefore,$l|\vec{r}| = 6, m|\vec{r}| = -3, n|\vec{r}| = 2$ $(i)$.
Squaring and adding these equations:
$(l|\vec{r}|)^2 + (m|\vec{r}|)^2 + (n|\vec{r}|)^2 = 6^2 + (-3)^2 + 2^2$
$|\vec{r}|^2(l^2 + m^2 + n^2) = 36 + 9 + 4$
Since $l^2 + m^2 + n^2 = 1$,we have $|\vec{r}|^2 = 49$,which implies $|\vec{r}| = 7$.
Substituting $|\vec{r}| = 7$ into equation $(i)$:
$l = \frac{6}{7}, m = -\frac{3}{7}, n = \frac{2}{7}$.

(B) Let the angles made by the line with the positive directions of the $x$,$y$,and $z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = \frac{\pi}{4}$ and $\beta = \frac{\pi}{4}$.
The direction cosines of the line 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 $\cos^2(\frac{\pi}{4}) + \cos^2(\frac{\pi}{4}) + \cos^2 \gamma = 1$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{1}{2} + \frac{1}{2} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$,which implies $\cos \gamma = 0$.
Therefore,$\gamma = \frac{\pi}{2}$.

(D) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.
Given $\alpha = 45^{\circ}$ and $\beta = 120^{\circ}$.
The relation between direction cosines is $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.
Substituting the values: $\cos^{2} 45^{\circ} + \cos^{2} 120^{\circ} + \cos^{2} \gamma = 1$.
$\left(\frac{1}{\sqrt{2}}\right)^{2} + \left(-\frac{1}{2}\right)^{2} + \cos^{2} \gamma = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^{2} \gamma = 1$.
$\frac{3}{4} + \cos^{2} \gamma = 1$.
$\cos^{2} \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $\theta = \gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$.
Therefore,$\gamma = 60^{\circ} = \frac{\pi}{3}$.

(D) Let the angles the line makes with the positive directions of the $x$,$y$,and $z$-axes be $\alpha, \beta$,and $\gamma$ respectively.
Given that $\alpha = 45^\circ$ and $\beta = 45^\circ$.
Let the direction cosines of the line be $l, m, n$.
Then $l = \cos \alpha = \cos 45^\circ = \frac{1}{\sqrt{2}}$,
$m = \cos \beta = \cos 45^\circ = \frac{1}{\sqrt{2}}$,
$n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{1}{2} + \frac{1}{2} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$.
$\cos \gamma = 0$.
Therefore,$\gamma = 90^\circ$.

(B) Let the side length of the cube be $a$.
Since the $3$ concurrent edges are along the coordinate axes,the origin is at $(0, 0, 0)$ and the opposite corner is at $(a, a, a)$.
The direction ratios of the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
Substituting the coordinates,we get $(a - 0, a - 0, a - 0) = (a, a, a)$.
Since direction ratios are proportional,we can divide by $a$ to get $(1, 1, 1)$.
Thus,the direction ratios are $1, 1, 1$.

(C) Let the direction cosines of the line be $(\cos \theta, \cos \beta, \cos \theta)$.
Since the sum of the squares of the direction cosines is $1$,we have:
$\cos^2 \theta + \cos^2 \beta + \cos^2 \theta = 1$
$2\cos^2 \theta + \cos^2 \beta = 1$
Using the identity $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get:
$2(1 - \sin^2 \theta) + (1 - \sin^2 \beta) = 1$
$3 - 2\sin^2 \theta - \sin^2 \beta = 1$
Given $\sin^2 \beta = 3\sin^2 \theta$,substitute this into the equation:
$3 - 2\sin^2 \theta - 3\sin^2 \theta = 1$
$3 - 5\sin^2 \theta = 1$
$5\sin^2 \theta = 2$
$\sin^2 \theta = \frac{2}{5}$
Now,calculate $\cos^2 \theta$:
$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{2}{5} = \frac{3}{5}$