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

(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.
From $(1)$,$m = -2l - 2n$.
Substitute $m$ into $(2)$: $3l^2 + 5(-2l - 2n)^2 - 11n^2 = 0$.
$3l^2 + 5(4l^2 + 8ln + 4n^2) - 11n^2 = 0$.
$3l^2 + 20l^2 + 40ln + 20n^2 - 11n^2 = 0$.
$23l^2 + 40ln + 9n^2 = 0$.
Divide by $n^2$: $23(\frac{l}{n})^2 + 40(\frac{l}{n}) + 9 = 0$.
Let $x = \frac{l}{n}$. Then $23x^2 + 40x + 9 = 0$.
Let the roots be $x_1 = \frac{l_1}{n_1}$ and $x_2 = \frac{l_2}{n_2}$.
Then $x_1 x_2 = \frac{l_1 l_2}{n_1 n_2} = \frac{9}{23}$.
Similarly,substituting $l = -\frac{m+2n}{2}$ into $(2)$ gives $3(\frac{m+2n}{2})^2 + 5m^2 - 11n^2 = 0$,which leads to $3(m^2 + 4mn + 4n^2) + 20m^2 - 44n^2 = 0$,so $23m^2 + 12mn - 32n^2 = 0$.
Dividing by $n^2$,$23(\frac{m}{n})^2 + 12(\frac{m}{n}) - 32 = 0$.
Let $y_1 = \frac{m_1}{n_1}$ and $y_2 = \frac{m_2}{n_2}$. Then $y_1 y_2 = \frac{m_1 m_2}{n_1 n_2} = -\frac{32}{23}$.
For two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Since $l_i^2 + m_i^2 + n_i^2 = 1$,we use the relation $l_1 l_2 + m_1 m_2 + n_1 n_2 = n_1 n_2 (x_1 x_2 + y_1 y_2 + 1) = n_1 n_2 (\frac{9}{23} - \frac{32}{23} + 1) = n_1 n_2 (\frac{-23}{23} + 1) = 0$.
Thus,$\cos \theta = 0$,which means $\theta = \frac{\pi}{2}$.

(B) Let the direction cosines of the line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.
The four diagonals of a cube are along the vectors $(\pm 1, \pm 1, \pm 1)$.
The unit vectors along the four diagonals are $\vec{d_1} = \frac{1}{\sqrt{3}}(1, 1, 1)$,$\vec{d_2} = \frac{1}{\sqrt{3}}(-1, 1, 1)$,$\vec{d_3} = \frac{1}{\sqrt{3}}(1, -1, 1)$,and $\vec{d_4} = \frac{1}{\sqrt{3}}(1, 1, -1)$.
The cosine of the angle $\theta$ between the line and a diagonal vector $\vec{d}$ is given by $|l \cdot d_x + m \cdot d_y + n \cdot d_z|$.
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 these,we get $\cos^2 \alpha = \frac{1}{3}(l+m+n)^2$,$\cos^2 \beta = \frac{1}{3}(-l+m+n)^2$,$\cos^2 \gamma = \frac{1}{3}(l-m+n)^2$,and $\cos^2 \delta = \frac{1}{3}(l+m-n)^2$.
Summing these: $\sum \cos^2 \alpha = \frac{1}{3} [ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 ]$.
Expanding the squares: $\sum \cos^2 \alpha = \frac{1}{3} [ 4(l^2+m^2+n^2) ] = \frac{4}{3}(1) = \frac{4}{3}$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sum \sin^2 \alpha = 4 - \sum \cos^2 \alpha = 4 - \frac{4}{3} = \frac{8}{3}$.

(C) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines satisfy $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given expression: $E = \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we have:
$E = (\cos^2 \alpha - \sin^2 \alpha) + (\cos^2 \beta - \sin^2 \beta) + (\cos^2 \gamma - \sin^2 \gamma) + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Simplifying the expression:
$E = \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + \cos^2 \gamma - \sin^2 \gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Canceling the $\sin^2$ terms:
$E = \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$.
Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,the value of the expression is $1$.

(C) Let the coordinates of point $P$ be $(u, v, w)$.
Since $P$ is equidistant from $O(0, 0, 0), A(3, 0, 0), B(0, 6, 0),$ and $C(0, 0, 9)$,we have $PO^2 = PA^2 = PB^2 = PC^2$.
$PO^2 = u^2 + v^2 + w^2$.
$PA^2 = (u-3)^2 + v^2 + w^2 = u^2 - 6u + 9 + v^2 + w^2$.
Equating $PO^2 = PA^2$,we get $u^2 = u^2 - 6u + 9$ $\Rightarrow 6u = 9$ $\Rightarrow u = \frac{3}{2}$.
$PB^2 = u^2 + (v-6)^2 + w^2 = u^2 + v^2 - 12v + 36 + w^2$.
Equating $PO^2 = PB^2$,we get $v^2 = v^2 - 12v + 36$ $\Rightarrow 12v = 36$ $\Rightarrow v = 3$.
$PC^2 = u^2 + v^2 + (w-9)^2 = u^2 + v^2 + w^2 - 18w + 81$.
Equating $PO^2 = PC^2$,we get $w^2 = w^2 - 18w + 81$ $\Rightarrow 18w = 81$ $\Rightarrow w = \frac{81}{18} = \frac{9}{2}$.
Thus,$P = \left(\frac{3}{2}, 3, \frac{9}{2}\right)$.
Given $Q = (2, 5, 8)$,point $R$ divides $PQ$ in the ratio $3:2$. Using the section formula $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$:
$R = \left(\frac{3(2) + 2(\frac{3}{2})}{3+2}, \frac{3(5) + 2(3)}{3+2}, \frac{3(8) + 2(\frac{9}{2})}{3+2}\right)$
$R = \left(\frac{6+3}{5}, \frac{15+6}{5}, \frac{24+9}{5}\right) = \left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$.

(A) Let the points be $A(3, 4, 5)$,$B(4, 6, 3)$,$C(-1, 2, 4)$,and $D(1, 0, 5)$.
First,we find the vector $\overrightarrow{AB}$ representing the line segment joining $A$ and $B$:
$\overrightarrow{AB} = (4-3)\hat{i} + (6-4)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$.
Next,we find the vector $\overrightarrow{CD}$ representing the line joining $C$ and $D$:
$\overrightarrow{CD} = (1 - (-1))\hat{i} + (0-2)\hat{j} + (5-4)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.
The length of the projection of vector $\overrightarrow{AB}$ on vector $\overrightarrow{CD}$ is given by the formula $\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|} \right|$.
Calculate the dot product $\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(2) + (2)(-2) + (-2)(1) = 2 - 4 - 2 = -4$.
Calculate the magnitude of $\overrightarrow{CD} = |\overrightarrow{CD}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Therefore,the length of the projection is $\left| \frac{-4}{3} \right| = \frac{4}{3}$.

(D) Since $OA$ is equally inclined to the axes $OX, OY$ and $OZ$,the direction cosines are equal. Let the coordinates of $A$ be $(a, a, a)$.
Given that the distance of $A$ from the origin $O(0,0,0)$ is $\sqrt{3}$ units.
Therefore,$\sqrt{(a-0)^2 + (a-0)^2 + (a-0)^2} = \sqrt{3}$.
$\sqrt{3a^2} = \sqrt{3}$.
$|a|\sqrt{3} = \sqrt{3}$.
$|a| = 1$,which implies $a = 1$ or $a = -1$.
Thus,the coordinates of $A$ are $(1, 1, 1)$ or $(-1, -1, -1)$.
Comparing this with the given options,the correct coordinate is $(1, 1, 1)$.

(A) Given equations are $2l+m-n=0$ $(1)$ and $l^2-2m^2+n^2=0$ $(2)$.
From $(1)$,$n = 2l+m$.
Substitute $n$ into $(2)$: $l^2 - 2m^2 + (2l+m)^2 = 0$.
$l^2 - 2m^2 + 4l^2 + 4lm + m^2 = 0$.
$5l^2 + 4lm - m^2 = 0$.
Divide by $m^2$: $5(l/m)^2 + 4(l/m) - 1 = 0$.
Let $x = l/m$,then $5x^2 + 4x - 1 = 0$.
$(5x-1)(x+1) = 0$,so $x = 1/5$ or $x = -1$.
Case $1$: $l/m = 1/5 \implies m = 5l$. Then $n = 2l + 5l = 7l$. Direction ratios are $(l, 5l, 7l)$ or $(1, 5, 7)$.
Case $2$: $l/m = -1 \implies m = -l$. Then $n = 2l - l = l$. Direction ratios are $(l, -l, l)$ or $(1, -1, 1)$.
Let the two lines have direction ratios $\vec{a} = (1, 5, 7)$ and $\vec{b} = (1, -1, 1)$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(1) + (5)(-1) + (7)(1)|}{\sqrt{1^2+5^2+7^2} \sqrt{1^2+(-1)^2+1^2}} = \frac{|1-5+7|}{\sqrt{1+25+49} \sqrt{3}} = \frac{3}{\sqrt{75} \sqrt{3}} = \frac{3}{\sqrt{225}} = \frac{3}{15} = \frac{1}{5}$.

(A) Let the direction angles of the line be $\alpha = 60^{\circ}$,$\beta = 45^{\circ}$,and $\gamma = \theta$.
We know that the sum of the squares of the direction cosines of a line is $1$,which is given by the formula $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values,we get $\cos^2 60^{\circ} + \cos^2 45^{\circ} + \cos^2 \theta = 1$.
Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \theta = 1$.
$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$.
$\frac{3}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $\theta$ is an acute angle,$\cos \theta = \frac{1}{2}$.
Thus,$\theta = 60^{\circ}$.
Therefore,$\tan \theta = \tan 60^{\circ} = \sqrt{3}$.

(B) Given: $l-m+n=0$ $\Rightarrow m=l+n$ $(i)$
Substituting $m=l+n$ into $2lm-3mn+nl=0$:
$2l(l+n)-3(l+n)n+nl=0$
$2l^2+2ln-3ln-3n^2+nl=0$
$2l^2-3n^2=0$
$l^2 = \frac{3}{2}n^2$
Let $n=1$,then $l^2 = \frac{3}{2} \Rightarrow l = \pm \sqrt{\frac{3}{2}}$.
For $l_1 = \sqrt{\frac{3}{2}}$,$m_1 = \sqrt{\frac{3}{2}}+1$.
For $l_2 = -\sqrt{\frac{3}{2}}$,$m_2 = -\sqrt{\frac{3}{2}}+1$.
The direction ratios of the two lines are $\vec{a} = (\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}+1, 1)$ and $\vec{b} = (-\sqrt{\frac{3}{2}}, 1-\sqrt{\frac{3}{2}}, 1)$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|-\frac{3}{2} + (1-\frac{3}{2}) + 1|}{\sqrt{\frac{3}{2} + (\frac{3}{2}+1+2\sqrt{\frac{3}{2}}) + 1} \cdot \sqrt{\frac{3}{2} + (1+\frac{3}{2}-2\sqrt{\frac{3}{2}}) + 1}}$
$\cos \theta = \frac{|-\frac{3}{2} - \frac{1}{2} + 1|}{\sqrt{4+2\sqrt{\frac{3}{2}}} \cdot \sqrt{4-2\sqrt{\frac{3}{2}}}} = \frac{|-1|}{\sqrt{16 - 4(\frac{3}{2})}} = \frac{1}{\sqrt{16-6}} = \frac{1}{\sqrt{10}}$.
Wait,re-evaluating the dot product: $\vec{a} \cdot \vec{b} = -\frac{3}{2} + (1 - \frac{3}{2}) + 1 = -\frac{3}{2} - \frac{1}{2} + 1 = -1$.
$|\vec{a}|^2 = \frac{3}{2} + 1 + \frac{3}{2} + 2\sqrt{\frac{3}{2}} + 1 = 4 + 2\sqrt{\frac{3}{2}}$.
$|\vec{b}|^2 = \frac{3}{2} + 1 + \frac{3}{2} - 2\sqrt{\frac{3}{2}} + 1 = 4 - 2\sqrt{\frac{3}{2}}$.
$|\vec{a}||\vec{b}| = \sqrt{16 - 4(\frac{3}{2})} = \sqrt{16-6} = \sqrt{10}$.
Thus $\cos \theta = \frac{1}{\sqrt{10}}$.
Given the options,the calculation yields $\frac{1}{\sqrt{10}}$. If we re-check the original equation $2lm-3mn+nl=0$,substituting $m=l+n$ gives $2l^2+2ln-3ln-3n^2+nl=0 \Rightarrow 2l^2-3n^2=0$. The result is $\frac{1}{\sqrt{10}}$.

(A) Given the relations for the direction cosines $(l, m, n)$ of the two lines:
$l+m+n=0$ and $lm=0$
From $lm=0$,we have two cases: $l=0$ or $m=0$.
Case $1$: If $l=0$,then $0+m+n=0 \Rightarrow n=-m$. The direction ratios are $(0, m, -m)$,which simplifies to $(0, 1, -1)$.
Case $2$: If $m=0$,then $l+0+n=0 \Rightarrow n=-l$. The direction ratios are $(l, 0, -l)$,which simplifies to $(1, 0, -1)$.
Let the direction vectors be $\vec{a} = (0, 1, -1)$ and $\vec{b} = (1, 0, -1)$.
The cosine of the angle $\theta$ between the two lines is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(0)(1) + (1)(0) + (-1)(-1)|}{\sqrt{0^2+1^2+(-1)^2} \cdot \sqrt{1^2+0^2+(-1)^2}}$
$\cos \theta = \frac{|0 + 0 + 1|}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(C) Given equations are $3l+2m+n=0$ ...$(1)$ and $2mn-3nl+5lm=0$ ...$(2)$.
From equation $(1)$,$n = -(3l+2m)$.
Substituting this into equation $(2)$:
$2m(-(3l+2m)) - 3l(-(3l+2m)) + 5lm = 0$
$-6ml - 4m^2 + 9l^2 + 6lm + 5lm = 0$
$9l^2 + 5lm - 4m^2 = 0$.
Dividing by $m^2$ (assuming $m \neq 0$),we get $9(\frac{l}{m})^2 + 5(\frac{l}{m}) - 4 = 0$.
Let $t = \frac{l}{m}$,then $9t^2 + 5t - 4 = 0$.
$(9t-4)(t+1) = 0$,so $t = \frac{4}{9}$ or $t = -1$.
Case $1$: $t = -1 \Rightarrow l = -m$. Then $n = -(3(-m)+2m) = m$. The direction ratios are $(-1, 1, 1)$.
Case $2$: $t = \frac{4}{9} \Rightarrow l = \frac{4}{9}m$. Then $n = -(3(\frac{4}{9}m)+2m) = -(\frac{4}{3}m+2m) = -\frac{10}{3}m$. The direction ratios are $(\frac{4}{9}, 1, -\frac{10}{3})$,which is equivalent to $(4, 9, -30)$.
Let the direction vectors be $\vec{a} = (-1, 1, 1)$ and $\vec{b} = (4, 9, -30)$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(-1)(4) + (1)(9) + (1)(-30)|}{\sqrt{(-1)^2+1^2+1^2} \sqrt{4^2+9^2+(-30)^2}}$
$= \frac{|-4 + 9 - 30|}{\sqrt{3} \sqrt{16+81+900}} = \frac{|-25|}{\sqrt{3} \sqrt{997}} = \frac{25}{\sqrt{2991}}$.

(C) Let the two lines be $L_1$ and $L_2$ with direction cosines $(\ell, m, n)$ and $(a, b, c)$ respectively.
For two lines with direction cosines $(\ell_1, m_1, n_1)$ and $(\ell_2, m_2, n_2)$,the direction ratios of the angle bisectors are given by $(\ell_1 \pm \ell_2, m_1 \pm m_2, n_1 \pm n_2)$.
Thus,for the given lines,the direction ratios of the bisectors are $(\ell \pm a, m \pm b, n \pm c)$.
Therefore,option $C$ is correct.

(A) Given the relations between the direction ratios $(a, b, c)$ of two lines:
$a - b + c = 0$ $(i)$
$a^2 - b^2 + 2c^2 = 0$ $(ii)$
From $(i)$,we have $c = b - a$.
Substituting this into $(ii)$:
$a^2 - b^2 + 2(b - a)^2 = 0$
$a^2 - b^2 + 2(b^2 - 2ab + a^2) = 0$
$a^2 - b^2 + 2b^2 - 4ab + 2a^2 = 0$
$3a^2 - 4ab + b^2 = 0$
$(3a - b)(a - b) = 0$
This gives two cases:
Case $1$: $b = 3a \implies a:b = 1:3$. Then $c = b - a = 3a - a = 2a$. So,the direction ratios are $(1, 3, 2)$.
Case $2$: $b = a \implies a:b = 1:1$. Then $c = b - a = a - a = 0$. So,the direction ratios are $(1, 1, 0)$.
The angle $\theta$ between the lines with direction ratios $(a_1, b_1, c_1) = (1, 3, 2)$ and $(a_2, b_2, c_2) = (1, 1, 0)$ is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
$\cos \theta = \frac{|(1)(1) + (3)(1) + (2)(0)|}{\sqrt{1^2 + 3^2 + 2^2} \sqrt{1^2 + 1^2 + 0^2}}$
$\cos \theta = \frac{|1 + 3 + 0|}{\sqrt{1 + 9 + 4} \sqrt{1 + 1 + 0}} = \frac{4}{\sqrt{14} \sqrt{2}} = \frac{4}{\sqrt{28}} = \frac{4}{2\sqrt{7}} = \frac{2}{\sqrt{7}}$.

(C) Let the points be $A(2, 3, -1)$ and $O(0, 0, 0)$.
The direction ratios of the line $OA$ are $(2-0, 3-0, -1-0) = (2, 3, -1)$.
The distance $OA$ is $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{r}, \frac{b}{r}, \frac{c}{r}$,where $(a, b, c)$ are direction ratios and $r$ is the distance.
Thus,$l = \frac{2}{\sqrt{14}}, m = \frac{3}{\sqrt{14}}, n = \frac{-1}{\sqrt{14}}$.
Alternatively,for the line $AO$,the direction ratios are $(0-2, 0-3, 0-(-1)) = (-2, -3, 1)$.
The direction cosines are $\frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}}$.
Comparing with the options,option $C$ is correct.

(C) Given the direction ratios of two lines are $(a_1, b_1, c_1) = (2, -1, 2)$ and $(a_2, b_2, c_2) = (K, -3, -5)$,and the angle between them is $\theta = 60^{\circ}$.
Using the formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the given values:
$\cos 60^{\circ} = \frac{|2K + (-1)(-3) + 2(-5)|}{\sqrt{2^2 + (-1)^2 + 2^2} \sqrt{K^2 + (-3)^2 + (-5)^2}}$
$\frac{1}{2} = \frac{|2K + 3 - 10|}{\sqrt{4 + 1 + 4} \sqrt{K^2 + 9 + 25}}$
$\frac{1}{2} = \frac{|2K - 7|}{3 \sqrt{K^2 + 34}}$
$3 \sqrt{K^2 + 34} = 2 |2K - 7|$
Squaring both sides:
$9(K^2 + 34) = 4(2K - 7)^2$
$9K^2 + 306 = 4(4K^2 - 28K + 49)$
$9K^2 + 306 = 16K^2 - 112K + 196$
$7K^2 - 112K - 110 = 0$

(C) The line $OP$ passes through the origin $O(0, 0, 0)$ and has direction ratios $3, -2, 6$.
Thus,the coordinates of any point $P$ on this line can be written as $(3\lambda, -2\lambda, 6\lambda)$ for some scalar $\lambda$.
The distance of point $P$ from the origin $O$ is given by $|OP| = \sqrt{(3\lambda)^2 + (-2\lambda)^2 + (6\lambda)^2}$.
Given that $|OP| = 63$,we have:
$\sqrt{9\lambda^2 + 4\lambda^2 + 36\lambda^2} = 63$
$\sqrt{49\lambda^2} = 63$
$7|\lambda| = 63$
$|\lambda| = 9$
Taking $\lambda = 9$,the coordinates of $P$ are $(3(9), -2(9), 6(9)) = (27, -18, 54)$.
Taking $\lambda = -9$,the coordinates of $P$ are $(-27, 18, -54)$.
Comparing with the given options,the correct coordinates are $(27, -18, 54)$.

(A) Let the vector be $\vec{a} = \hat{i} + \hat{j} - 2\hat{k}$.
Its direction ratios are $(a, b, c) = (1, 1, -2)$.
The magnitude of the vector is $|\vec{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
The direction cosines $(l, m, n)$ are given by $\left(\frac{a}{|\vec{a}|}, \frac{b}{|\vec{a}|}, \frac{c}{|\vec{a}|}\right)$.
Substituting the values,we get $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\right)$.

(B) We know that the direction cosines are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
Also,the sum of squares of direction cosines is $l^2 + m^2 + n^2 = 1$.
We want to maximize the expression $S = lm + mn + nl$.
Using the identity $(l + m + n)^2 = l^2 + m^2 + n^2 + 2(lm + mn + nl)$,we have $2(lm + mn + nl) = (l + m + n)^2 - (l^2 + m^2 + n^2) = (l + m + n)^2 - 1$.
To maximize $lm + mn + nl$,we must maximize $(l + m + n)^2$.
By the Cauchy-Schwarz inequality,$(l + m + n)^2 \leq (1^2 + 1^2 + 1^2)(l^2 + m^2 + n^2) = 3(1) = 3$.
The equality holds when $l = m = n$.
Since $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$,this implies $\cos \alpha = \cos \beta = \cos \gamma$,which means $\alpha = \beta = \gamma$ (given the angles are between $0$ and $\pi$).
Thus,the maximum value is attained when $\alpha = \beta = \gamma$.

(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.
From $l+m+n=0$,we have $l=-(m+n)$.
Substituting this into $l^2=m^2+n^2$,we get $(-(m+n))^2 = m^2+n^2$.
$m^2+n^2+2mn = m^2+n^2$,which implies $2mn=0$,so $mn=0$.
This gives two cases:
Case $I$: $m=0$. Then $l=-n$. Let the direction ratios be $(k, 0, -k)$. The unit vector is $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
Case $II$: $n=0$. Then $l=-m$. Let the direction ratios be $(k, -k, 0)$. The unit vector is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
Let $\vec{a} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = |\vec{a} \cdot \vec{b}|$.
$\cos \theta = |(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0)| = |\frac{1}{2} + 0 + 0| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) Given equations are $3l + m + 5n = 0$ and $6mn - 2nl + 5lm = 0$.
From the first equation,$m = -3l - 5n$.
Substituting this into the second equation:
$6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0$
$-18nl - 30n^2 - 2nl - 15l^2 - 25nl = 0$
$-15l^2 - 45nl - 30n^2 = 0$
Dividing by $-15$:
$l^2 + 3nl + 2n^2 = 0$
$(l + n)(l + 2n) = 0$
This gives two cases:
Case $1$: $l = -n$. Substituting into $m = -3l - 5n$,we get $m = -3(-n) - 5n = 3n - 5n = -2n$.
The direction ratios are $(-n, -2n, n)$,which simplifies to $(1, 2, -1)$.
Case $2$: $l = -2n$. Substituting into $m = -3l - 5n$,we get $m = -3(-2n) - 5n = 6n - 5n = n$.
The direction ratios are $(-2n, n, n)$,which simplifies to $(-2, 1, 1)$.
Let the direction ratios be $\vec{a} = (1, 2, -1)$ and $\vec{b} = (-2, 1, 1)$.
The cosine of the angle $\theta$ is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(-2) + (2)(1) + (-1)(1)|}{\sqrt{1^2 + 2^2 + (-1)^2} \sqrt{(-2)^2 + 1^2 + 1^2}}$
$\cos \theta = \frac{|-2 + 2 - 1|}{\sqrt{6} \sqrt{6}} = \frac{|-1|}{6} = \frac{1}{6}$.

(D) Given the equations:
$ab + bc + ca = 0$ --- $(1)$
$6ab + 9bc + 8ca = 0$ --- $(2)$
Multiply equation $(1)$ by $6$:
$6ab + 6bc + 6ca = 0$ --- $(3)$
Subtracting $(3)$ from $(2)$:
$(6ab + 9bc + 8ca) - (6ab + 6bc + 6ca) = 0$
$3bc + 2ca = 0 \Rightarrow c(3b + 2a) = 0$.
Since $c \neq 0$ (otherwise $a=b=0$,which is not possible for direction ratios),we get $2a = -3b$,or $a/(-3) = b/2$.
Substitute $a = -3k$ and $b = 2k$ into equation $(1)$:
$(-3k)(2k) + (2k)c + c(-3k) = 0$
$-6k^2 - kc = 0 \Rightarrow c = -6k$.
Thus,the direction ratios are proportional to $(-3k, 2k, -6k)$,i.e.,$(-3, 2, -6)$.
The magnitude is $\sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Therefore,the direction cosines are $\frac{-3}{7}, \frac{2}{7}, \frac{-6}{7}$.

(A) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.
Given that $E, F, G$ are mid-points of $AB, BC, CA$ respectively:
$\frac{x_1+x_2}{2}=1, \frac{y_1+y_2}{2}=0, \frac{z_1+z_2}{2}=0$ $(1)$
$\frac{x_2+x_3}{2}=0, \frac{y_2+y_3}{2}=2, \frac{z_2+z_3}{2}=0$ $(2)$
$\frac{x_3+x_1}{2}=0, \frac{y_3+y_1}{2}=0, \frac{z_3+z_1}{2}=3$ $(3)$
Solving these,we get $A(1, -2, 3), B(1, 2, -3), C(-1, 2, 3)$.
Direction ratios of $AF$ (where $F$ is $(0, 2, 0)$): $a_1 = 0-1 = -1, b_1 = 2-(-2) = 4, c_1 = 0-3 = -3$. Thus $a_1^2+b_1^2+c_1^2 = (-1)^2+4^2+(-3)^2 = 1+16+9 = 26$.
Direction ratios of $BG$ (where $G$ is $(0, 0, 3)$): $a_2 = 0-1 = -1, b_2 = 0-2 = -2, c_2 = 3-(-3) = 6$. Thus $a_2^2+b_2^2+c_2^2 = (-1)^2+(-2)^2+6^2 = 1+4+36 = 41$.
Therefore,$\frac{a_1^2+b_1^2+c_1^2}{a_2^2+b_2^2+c_2^2} = \frac{26}{41}$.

(C) For two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$,the lines are parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given direction ratios for $L_1$ are $(2, 5, 7)$ and for $L_2$ are $(\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}})$.
Calculating the ratios: $\frac{2}{4/\sqrt{19}} = \frac{\sqrt{19}}{2}$,$\frac{5}{10/\sqrt{19}} = \frac{\sqrt{19}}{2}$,and $\frac{7}{14/\sqrt{19}} = \frac{\sqrt{19}}{2}$.
Since all ratios are equal,the lines $L_1$ and $L_2$ are parallel. Thus,$(A)$ is true.
The condition $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ is the condition for two lines to be perpendicular,not parallel. Thus,$(R)$ is false.
Therefore,$(A)$ is true but $(R)$ is false.

(B) Given equations are $a+b+c=0$ and $2ab+2ac-bc=0$.
Substituting $a=-(b+c)$ into the second equation:
$-2(b+c)b - 2(b+c)c - bc = 0$
$-2b^2 - 2bc - 2bc - 2c^2 - bc = 0$
$-2b^2 - 5bc - 2c^2 = 0$
$2b^2 + 5bc + 2c^2 = 0$
$(2b+c)(b+2c) = 0$.
Case $1$: $b = -2c$. Then $a = -(-2c+c) = c$. Direction ratios are $(1, -2, 1)$.
Case $2$: $b = -c/2$. Then $a = -(-c/2+c) = -c/2$. Direction ratios are $(-1/2, -1/2, 1)$,which is equivalent to $(1, 1, -2)$.
Let the direction ratios be $(a_1, b_1, c_1) = (1, -2, 1)$ and $(a_2, b_2, c_2) = (1, 1, -2)$.
Using the formula $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$:
$\cos \theta = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1+4+1} \sqrt{1+1+4}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$.
Since we need the obtuse angle,$\cos \theta = -1/2$,which gives $\theta = \frac{2 \pi}{3}$.

(D) Given equations are $a+2b+c=0$ $(i)$ and $11bc+6ca-14ab=0$ (ii).
From $(i)$,$b = \frac{-(a+c)}{2}$.
Substituting this into (ii):
$11\left(\frac{-(a+c)}{2}\right)c + 6ac - 14a\left(\frac{-(a+c)}{2}\right) = 0$
$\Rightarrow -\frac{11}{2}ac - \frac{11}{2}c^2 + 6ac + 7a^2 + 7ac = 0$
Multiplying by $2$: $-11ac - 11c^2 + 12ac + 14a^2 + 14ac = 0$
$\Rightarrow 14a^2 + 15ac - 11c^2 = 0$
Factoring the quadratic: $(7a+11c)(2a-c) = 0$.
Case $1$: $2a = c \Rightarrow a = 1, c = 2$. Then $b = \frac{-(1+2)}{2} = -1.5$. To avoid fractions,let $a=2, c=4, b=-3$. Direction ratios: $(2, -3, 4)$.
Case $2$: $7a = -11c \Rightarrow a = -11, c = 7$. Then $b = \frac{-(-11+7)}{2} = 2$. Direction ratios: $(-11, 2, 7)$.
Let the direction ratios be $\vec{n_1} = (2, -3, 4)$ and $\vec{n_2} = (-11, 2, 7)$.
The dot product $\vec{n_1} \cdot \vec{n_2} = (2)(-11) + (-3)(2) + (4)(7) = -22 - 6 + 28 = 0$.
Since the dot product is $0$,the lines are perpendicular.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(A) Given equations for direction cosines are $a l + b m + c n = 0$ $(1)$ and $m n + n l + l m = 0$ $(2)$.
From $(1)$,we have $n = -\frac{a l + b m}{c}$.
Substituting this into $(2)$:
$m \left( -\frac{a l + b m}{c} \right) + l \left( -\frac{a l + b m}{c} \right) + l m = 0$.
Multiplying by $-c$:
$m(a l + b m) + l(a l + b m) - c l m = 0$.
$a l^2 + b m^2 + a l m + b l m - c l m = 0$.
$a l^2 + (a + b - c) l m + b m^2 = 0$.
Dividing by $m^2$,we get $a \left( \frac{l}{m} \right)^2 + (a + b - c) \left( \frac{l}{m} \right) + b = 0$.
Let the two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$. The roots of the quadratic are $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$.
Thus,$\frac{l_1 l_2}{m_1 m_2} = \frac{b}{a}$,which implies $\frac{l_1 l_2}{b} = \frac{m_1 m_2}{a}$.
By symmetry,$\frac{l_1 l_2}{1/a} = \frac{m_1 m_2}{1/b} = \frac{n_1 n_2}{1/c} = k$.
The lines are perpendicular if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Substituting the values: $k \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 0$.
Since $k \neq 0$,the condition is $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$.

(C) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ --- $(i)$
$l^2+m^2-n^2=0$ --- $(ii)$
From $(i)$,$n = -(l+m)$.
Substituting this into $(ii)$:
$l^2 + m^2 - (-(l+m))^2 = 0$
$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$
$-2lm = 0 \Rightarrow lm = 0$.
This implies either $l=0$ or $m=0$.
Case $1$: If $l=0$,then from $(i)$,$m+n=0 \Rightarrow m=-n$. Let $m=1$,then $n=-1$. The direction ratios are $(0, 1, -1)$.
Case $2$: If $m=0$,then from $(i)$,$l+n=0 \Rightarrow l=-n$. Let $l=1$,then $n=-1$. The direction ratios are $(1, 0, -1)$.
Let the two 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,$\theta = \frac{\pi}{3}$.

(D) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0 \Rightarrow m=-(l+n) \quad (i)$
$2lm+2ln-mn=0 \quad (ii)$
Substituting $m=-(l+n)$ into $(ii)$:
$2l(-(l+n)) + 2ln - (-(l+n))n = 0$
$-2l^2 - 2ln + 2ln + ln + n^2 = 0$
$-2l^2 + ln + n^2 = 0$
$2l^2 - ln - n^2 = 0$
$(2l+n)(l-n) = 0$
This gives two cases:
Case $1$: $l=n$. From $(i)$,$m=-(n+n)=-2n$. Thus,$(l, m, n) = (n, -2n, n)$,which gives direction ratios $(1, -2, 1)$.
Case $2$: $2l=-n \Rightarrow l=-\frac{n}{2}$. From $(i)$,$m=-(-\frac{n}{2}+n) = -\frac{n}{2}$. Thus,$(l, m, n) = (-\frac{n}{2}, -\frac{n}{2}, n)$,which gives direction ratios $(1, 1, -2)$.
Let the direction ratios be $\vec{a} = (1, -2, 1)$ and $\vec{b} = (1, 1, -2)$.
The angle $\theta$ between the lines is given by:
$\cos \theta = \left| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right| = \left| \frac{(1)(1) + (-2)(1) + (1)(-2)}{\sqrt{1^2+(-2)^2+1^2} \sqrt{1^2+1^2+(-2)^2}} \right|$
$\cos \theta = \left| \frac{1 - 2 - 2}{\sqrt{6} \sqrt{6}} \right| = \left| \frac{-3}{6} \right| = \frac{1}{2}$
$\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^{\circ}$.

(A) Given equations are:
$l+3m+5n=0$ --- $(i)$
$5lm-2mn+6ln=0$ --- $(ii)$
From $(i)$,$l = -3m - 5n$.
Substituting this into $(ii)$:
$5(-3m-5n)m - 2mn + 6(-3m-5n)n = 0$
$-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$
$-15m^2 - 45mn - 30n^2 = 0$
Dividing by $-15$:
$m^2 + 3mn + 2n^2 = 0$
$(m+n)(m+2n) = 0$
Case $1$: $m = -n$. Then $l = -3(-n) - 5n = -2n$. Direction ratios are $(-2n, -n, n)$ or $(2, 1, -1)$.
Case $2$: $m = -2n$. Then $l = -3(-2n) - 5n = n$. Direction ratios are $(n, -2n, n)$ or $(1, -2, 1)$.
Let the direction ratios be $\vec{a} = (2, 1, -1)$ and $\vec{b} = (1, -2, 1)$.
The angle $\theta$ between the lines is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(2)(1) + (1)(-2) + (-1)(1)|}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+(-2)^2+1^2}}$
$\cos \theta = \frac{|2 - 2 - 1|}{\sqrt{6} \sqrt{6}} = \frac{|-1|}{6} = \frac{1}{6}$
Therefore,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(B) Given the direction cosines of two lines as $(l_1, m_1, n_1) = \left(-\frac{2}{\sqrt{21}}, \frac{C}{\sqrt{21}}, \frac{1}{\sqrt{21}}\right)$ and $(l_2, m_2, n_2) = \left(\frac{3}{\sqrt{54}}, \frac{3}{\sqrt{54}}, -\frac{6}{\sqrt{54}}\right)$.
Since the angle between the lines is $\frac{\pi}{2}$,the lines are perpendicular.
For perpendicular lines,the condition is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Substituting the values:
$\left(-\frac{2}{\sqrt{21}}\right) \left(\frac{3}{\sqrt{54}}\right) + \left(\frac{C}{\sqrt{21}}\right) \left(\frac{3}{\sqrt{54}}\right) + \left(\frac{1}{\sqrt{21}}\right) \left(-\frac{6}{\sqrt{54}}\right) = 0$
$\frac{-6}{\sqrt{21}\sqrt{54}} + \frac{3C}{\sqrt{21}\sqrt{54}} - \frac{6}{\sqrt{21}\sqrt{54}} = 0$
Multiplying by $\sqrt{21}\sqrt{54}$ on both sides:
$-6 + 3C - 6 = 0$
$3C - 12 = 0$
$3C = 12$
$C = 4$.

(D) Given the equations for direction cosines $l, m, n$:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2=0$
$l^2-5m^2+l^2+2lm+m^2=0$
$2l^2+2lm-4m^2=0$
$l^2+lm-2m^2=0$
$(l+2m)(l-m)=0$
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$,i.e.,$(-2, 1, 1)$.
Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.