Concurrency of three lines Questions in English

(D) Let the points be $A(3a, 0)$,$B(0, 3b)$,and $C(a, 2b)$.
To check if the points are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |3a(3b - 2b) + 0(2b - 0) + a(0 - 3b)|$
$\text{Area} = \frac{1}{2} |3a(b) + 0 - 3ab|$
$\text{Area} = \frac{1}{2} |3ab - 3ab| = 0$
Since the area of the triangle is $0$,the points are collinear.

(D) Let the points be $A(-a, -b)$,$B(a, b)$,and $C(a^2, ab)$.
Calculate the slopes of the line segments:
Slope of $AB = \frac{b - (-b)}{a - (-a)} = \frac{2b}{2a} = \frac{b}{a}$.
Slope of $BC = \frac{ab - b}{a^2 - a} = \frac{b(a - 1)}{a(a - 1)} = \frac{b}{a}$.
Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A, B$,and $C$ are collinear.

(A) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is zero.
The area of the triangle is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points:
$\frac{1}{2} |k(2k - (6 - 2k)) + (1 - k)((6 - 2k) - (2 - 2k)) + (-k - 4)((2 - 2k) - 2k)| = 0$
Simplifying the terms inside the brackets:
$k(4k - 6) + (1 - k)(4) + (-k - 4)(2 - 4k) = 0$
$4k^2 - 6k + 4 - 4k + (-2k + 4k^2 - 8 + 16k) = 0$
$4k^2 - 10k + 4 + 4k^2 + 14k - 8 = 0$
$8k^2 + 4k - 4 = 0$
Dividing by $4$:
$2k^2 + k - 1 = 0$
$(2k - 1)(k + 1) = 0$
Thus,$k = 1/2$ or $k = -1$.

(D) Let $A \equiv (x + 1, 2)$,$B \equiv (1, x + 2)$,and $C \equiv \left( \frac{1}{x + 1}, \frac{2}{x + 1} \right)$.
The points $A, B, C$ are collinear if the area of $\Delta ABC = 0$.
$\left| \begin{array}{ccc} x + 1 & 2 & 1 \\ 1 & x + 2 & 1 \\ \frac{1}{x + 1} & \frac{2}{x + 1} & 1 \end{array} \right| = 0$
Applying $R_1 \to R_1 - R_2$:
$\left| \begin{array}{ccc} x & -x & 0 \\ 1 & x + 2 & 1 \\ \frac{1}{x + 1} & \frac{2}{x + 1} & 1 \end{array} \right| = 0$
Applying $C_2 \to C_2 + C_1$:
$\left| \begin{array}{ccc} x & 0 & 0 \\ 1 & x + 3 & 1 \\ \frac{1}{x + 1} & \frac{3}{x + 1} & 1 \end{array} \right| = 0$
Expanding along $R_1$:
$x \left( (x + 3)(1) - (1)\left( \frac{3}{x + 1} \right) \right) = 0$
$x \left( x + 3 - \frac{3}{x + 1} \right) = 0$
$x \left( \frac{(x + 3)(x + 1) - 3}{x + 1} \right) = 0$
$x \left( \frac{x^2 + 4x + 3 - 3}{x + 1} \right) = 0$
$\frac{x(x^2 + 4x)}{x + 1} = 0$
$\frac{x^2(x + 4)}{x + 1} = 0$
Thus,$x = 0$ or $x = -4$.

(A) Step $1$: Find the point of intersection of the lines $x + 5y + 7 = 0$ and $3x + 2y - 5 = 0$.
Multiply the first equation by $3$: $3x + 15y + 21 = 0$.
Subtract the second equation from this: $(3x + 15y + 21) - (3x + 2y - 5) = 0$,which gives $13y + 26 = 0$,so $y = -2$.
Substitute $y = -2$ into $x + 5y + 7 = 0$: $x + 5(-2) + 7 = 0 \implies x - 10 + 7 = 0 \implies x = 3$.
The point of intersection is $(3, -2)$.
Step $2$: Find the slope of the line perpendicular to $7x + 2y - 5 = 0$.
The slope of $7x + 2y - 5 = 0$ is $m_1 = -\frac{7}{2}$.
The slope of the required line $m_2$ satisfies $m_1 \times m_2 = -1$,so $m_2 = \frac{2}{7}$.
Step $3$: Write the equation of the line using the point-slope form $y - y_1 = m(x - x_1)$.
$y - (-2) = \frac{2}{7}(x - 3) \implies 7(y + 2) = 2(x - 3) \implies 7y + 14 = 2x - 6$.
Rearranging gives $2x - 7y - 20 = 0$.

(A) Step $1$: Find the point of intersection of the lines $2x + y = 5$ and $x + 3y = -8$.
Multiply the first equation by $3$: $6x + 3y = 15$.
Subtract the second equation from this: $(6x + 3y) - (x + 3y) = 15 - (-8) \implies 5x = 23 \implies x = \frac{23}{5}$.
Substitute $x$ into $2x + y = 5$: $2(\frac{23}{5}) + y = 5 \implies y = 5 - \frac{46}{5} = -\frac{21}{5}$.
The point of intersection is $(\frac{23}{5}, -\frac{21}{5})$.
Step $2$: $A$ line parallel to $3x + 4y = 7$ is of the form $3x + 4y + k = 0$.
Since it passes through $(\frac{23}{5}, -\frac{21}{5})$,we have $3(\frac{23}{5}) + 4(-\frac{21}{5}) + k = 0$.
$\frac{69 - 84}{5} + k = 0 \implies -\frac{15}{5} + k = 0 \implies -3 + k = 0 \implies k = 3$.
Thus,the equation is $3x + 4y + 3 = 0$.

(C) Step $1$: Find the point of intersection of the lines $3x - 2y - 1 = 0$ and $x - 4y + 3 = 0$.
Multiply the second equation by $3$: $3x - 12y + 9 = 0$.
Subtract this from the first equation: $(3x - 2y - 1) - (3x - 12y + 9) = 0$ $\Rightarrow 10y - 10 = 0$ $\Rightarrow y = 1$.
Substitute $y = 1$ into $x - 4y + 3 = 0$: $x - 4(1) + 3 = 0$ $\Rightarrow x - 1 = 0$ $\Rightarrow x = 1$.
The point of intersection is $(1, 1)$.
Step $2$: Find the equation of the line passing through $(1, 1)$ and $(\pi, 0)$.
The slope $m = \frac{0 - 1}{\pi - 1} = \frac{-1}{\pi - 1}$.
The equation is $y - 0 = \frac{-1}{\pi - 1}(x - \pi)$.
$(\pi - 1)y = -x + \pi$ $\Rightarrow x + (\pi - 1)y = \pi$ $\Rightarrow x - \pi = -(\pi - 1)y$ $\Rightarrow x - \pi = -\pi y + y$ $\Rightarrow x - y = \pi - \pi y$ $\Rightarrow x - y = \pi(1 - y)$.

(B) Step $1$: Find the point of intersection of $4x - 3y = 1$ and $5x - 2y = 3$.
Multiplying the first equation by $2$ and the second by $3$:
$8x - 6y = 2$
$15x - 6y = 9$
Subtracting the equations: $(15x - 8x) = 9 - 2 \implies 7x = 7 \implies x = 1$.
Substituting $x = 1$ in $4(1) - 3y = 1 \implies 3y = 3 \implies y = 1$.
The point of intersection is $(1, 1)$.
Step $2$: The equation of a line parallel to $2y - 3x + 2 = 0$ is of the form $2y - 3x + k = 0$ or $3x - 2y = C$.
Step $3$: Since the line passes through $(1, 1)$,substitute these coordinates into $3x - 2y = C$:
$3(1) - 2(1) = C \implies C = 1$.
Thus,the required equation is $3x - 2y = 1$.

(B) To find the point of intersection of the lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$,we solve them simultaneously.
Multiply the first equation by $2$: $2x + 4y - 20 = 0$.
Subtract the second equation from this: $(2x + 4y - 20) - (2x + y + 5) = 0$,which simplifies to $3y - 25 = 0$,so $y = 25/3$.
Substitute $y = 25/3$ into $x + 2y - 10 = 0$: $x + 2(25/3) - 10 = 0 \implies x + 50/3 - 30/3 = 0 \implies x = -20/3$.
The point of intersection is $(-20/3, 25/3)$.
Testing the options,for $5x + 4y = 0$:
$5(-20/3) + 4(25/3) = -100/3 + 100/3 = 0$.
Thus,the line $5x + 4y = 0$ passes through the point of intersection.

(A) Given lines are $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$ and $\frac{x}{b} + \frac{y}{a} = 1$ $(2)$.
Adding $(1)$ and $(2)$: $x(\frac{1}{a} + \frac{1}{b}) + y(\frac{1}{a} + \frac{1}{b}) = 2 \implies (x + y)(\frac{a + b}{ab}) = 2 \implies x + y = \frac{2ab}{a + b}$.
Subtracting $(2)$ from $(1)$: $x(\frac{1}{a} - \frac{1}{b}) + y(\frac{1}{b} - \frac{1}{a}) = 0 \implies x(\frac{b - a}{ab}) - y(\frac{b - a}{ab}) = 0 \implies x = y$.
Substituting $x = y$ in $x + y = \frac{2ab}{a + b}$,we get $2x = \frac{2ab}{a + b} \implies x = \frac{ab}{a + b}$.
So,the intersection point is $P = (\frac{ab}{a + b}, \frac{ab}{a + b})$.
The line joins $P$ and $Q = (a, b)$. The slope $m = \frac{b - \frac{ab}{a + b}}{a - \frac{ab}{a + b}} = \frac{\frac{ab + b^2 - ab}{a + b}}{\frac{a^2 + ab - ab}{a + b}} = \frac{b^2}{a^2}$.
The equation is $y - b = \frac{b^2}{a^2}(x - a) \implies a^2y - a^2b = b^2x - ab^2 \implies a^2y - b^2x = a^2b - ab^2 = ab(a - b)$.

(C) Step $1$: Find the intersection point of the lines $x - 2y = 1$ and $x + 3y = 2$. Subtracting the first from the second gives $5y = 1$,so $y = \frac{1}{5}$. Substituting $y$ into the first equation,$x - 2(\frac{1}{5}) = 1$,which gives $x = 1 + \frac{2}{5} = \frac{7}{5}$. The intersection point is $(\frac{7}{5}, \frac{1}{5})$.
Step $2$: The required line is parallel to $3x + 4y = 0$,so its slope is $m = -\frac{3}{4}$.
Step $3$: Using the point-slope form $y - y_1 = m(x - x_1)$,we get $y - \frac{1}{5} = -\frac{3}{4}(x - \frac{7}{5})$.
Step $4$: Multiplying by $20$ to clear denominators: $20y - 4 = -15(x - \frac{7}{5}) \Rightarrow 20y - 4 = -15x + 21$.
Step $5$: Rearranging the terms gives $15x + 20y = 25$,which simplifies to $3x + 4y = 5$ or $3x + 4y - 5 = 0$.

(B) Step $1$: Find the point of intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we get $x = -\frac{1}{17}$ and $y = \frac{22}{17}$. The point is $(-\frac{1}{17}, \frac{22}{17})$.
Step $2$: The given line is $6x - 7y + 3 = 0$,which has a slope of $m_1 = \frac{6}{7}$.
Step $3$: The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$.
Step $4$: Using the point-slope form $y - y_1 = m(x - x_1)$,we have $y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$.
Step $5$: Multiplying by $102$,we get $102y - 132 = -119x - 7$,which simplifies to $119x + 102y = 125$.

(C) The equation of any line passing through the intersection of $L_1: 3x - y + 2 = 0$ and $L_2: 5x - 2y + 7 = 0$ is given by $L_1 + \lambda L_2 = 0$.
$(3x - y + 2) + \lambda (5x - 2y + 7) = 0$
$(3 + 5\lambda)x - (1 + 2\lambda)y + (2 + 7\lambda) = 0$
For a line to have an infinite slope,the coefficient of $y$ must be zero.
$1 + 2\lambda = 0 \implies \lambda = -1/2$
Substituting $\lambda = -1/2$ into the equation:
$(3x - y + 2) - \frac{1}{2}(5x - 2y + 7) = 0$
Multiply by $2$:
$2(3x - y + 2) - (5x - 2y + 7) = 0$
$6x - 2y + 4 - 5x + 2y - 7 = 0$
$x - 3 = 0 \implies x = 3$.

(A) The given equation is $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$.
Rearranging the terms with respect to $a$ and $b$,we get:
$a(x + y + 3) + b(-2x + 3y + 4) = 0$.
This represents a family of straight lines passing through the intersection of the lines $x + y + 3 = 0$ and $-2x + 3y + 4 = 0$.
Solving the system of equations:
$x + y = -3$ $(i)$
$-2x + 3y = -4$ (ii)
Multiplying $(i)$ by $2$: $2x + 2y = -6$.
Adding this to (ii): $(2x - 2x) + (2y + 3y) = -6 - 4$,which gives $5y = -10$,so $y = -2$.
Substituting $y = -2$ into $(i)$: $x - 2 = -3$,so $x = -1$.
Thus,the point of intersection is $(-1, -2)$.

(A) Given the condition $a + b + c = 0$,we can substitute $(b + c) = -a$,$(c + a) = -b$,and $(a + b) = -c$.
The equations of the lines become:
$ax - ay = p \implies x - y = \frac{p}{a}$
$bx - by = p \implies x - y = \frac{p}{b}$
$cx - cy = p \implies x - y = \frac{p}{c}$
Since $p \neq 0$ and $a, b, c$ are distinct (assuming non-zero values for the lines to be distinct),the slopes of all three lines are $1$.
Because the lines have the same slope but different $y$-intercepts (since $\frac{p}{a} \neq \frac{p}{b} \neq \frac{p}{c}$),they are parallel to each other.
Therefore,the lines do not intersect in the finite plane.

(D) Given lines are:
$1) \frac{x}{a} + \frac{y}{b} = 1$
$2) \frac{x}{b} + \frac{y}{a} = 1$
Adding the two equations:
$(\frac{x}{a} + \frac{x}{b}) + (\frac{y}{b} + \frac{y}{a}) = 2$
$x(\frac{a+b}{ab}) + y(\frac{a+b}{ab}) = 2$
$(x + y)(\frac{a+b}{ab}) = 2 \implies x + y = \frac{2ab}{a+b}$
Subtracting the two equations:
$(\frac{x}{a} - \frac{x}{b}) + (\frac{y}{b} - \frac{y}{a}) = 0$
$x(\frac{b-a}{ab}) - y(\frac{b-a}{ab}) = 0$
$(x - y)(\frac{b-a}{ab}) = 0 \implies x = y$
Substituting $x = y$ into $x + y = \frac{2ab}{a+b}$,we get $2x = \frac{2ab}{a+b} \implies x = \frac{ab}{a+b}$.
Thus,the intersection point is $(\frac{ab}{a+b}, \frac{ab}{a+b})$.
Checking option $(a)$: $x - y = \frac{ab}{a+b} - \frac{ab}{a+b} = 0$. (Correct)
Checking option $(b)$: $(x + y)(a + b) = (\frac{ab}{a+b} + \frac{ab}{a+b})(a + b) = (\frac{2ab}{a+b})(a + b) = 2ab$. (Correct)
Checking option $(c)$: $(lx + my)(a + b) = (l(\frac{ab}{a+b}) + m(\frac{ab}{a+b}))(a + b) = (l+m)(\frac{ab}{a+b})(a+b) = (l+m)ab$. (Correct)
Therefore,all options are correct.

(D) Two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are identical if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$.
For the given equations:
$\frac{b^3 - c^3}{b - c} = \frac{c^3 - a^3}{c - a} = \frac{a^3 - b^3}{a - b} = k$
Using the identity $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$,we get:
$b^2 + bc + c^2 = c^2 + ca + a^2 = a^2 + ab + b^2 = k$
Case $1$: If $a = b = c$,the equations become $0 = 0$,which represents the same line.
Case $2$: If $a, b, c$ are not all equal,then $b^2 + bc + c^2 = c^2 + ca + a^2 \implies b^2 - a^2 + c(b - a) = 0 \implies (b - a)(b + a + c) = 0$.
This implies $b = a$ or $a + b + c = 0$.
Since the condition must hold for any of the variables being equal,if $a=b$,$b=c$,or $c=a$,the lines are identical. Thus,all the above are correct.

(C) To find the intersection point,we solve the system of equations:
$ax + 2by = -3b$ $(1)$
$bx - 2ay = 3a$ $(2)$
Multiply $(1)$ by $a$ and $(2)$ by $b$:
$a^2x + 2aby = -3ab$
$b^2x - 2aby = 3ab$
Adding these equations: $(a^2 + b^2)x = 0$. Since $(a, b) \ne (0, 0)$,$a^2 + b^2 \ne 0$,so $x = 0$.
Substitute $x = 0$ into $(1)$: $2by = -3b$. Since $b$ could be $0$,we check the case $b=0$. If $b=0$,then $ax=0 \Rightarrow x=0$ and $-2ay=3a \Rightarrow y=-3/2$. If $b \ne 0$,$y = -3/2$.
Thus,the intersection point is $(0, -3/2)$.
$A$ line parallel to the $x$-axis has the form $y = k$. Since it passes through $(0, -3/2)$,the line is $y = -3/2$.
This line is $3/2$ units below the $x$-axis.

(C) The given lines are:
$ax + by + c = 0$ $(i)$
$bx + cy + a = 0$ $(ii)$
$cx + ay + b = 0$ $(iii)$
Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$
Expanding the determinant:
$a(cb - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$
$abc - a^3 - b^3 + abc + abc - c^3 = 0$
$3abc - (a^3 + b^3 + c^3) = 0$
Therefore,$a^3 + b^3 + c^3 - 3abc = 0$.

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$
Rewriting the given lines as $m_1x - y + c_1 = 0$,$m_2x - y + c_2 = 0$,and $m_3x - y + c_3 = 0$,the condition for concurrency is:
$\begin{vmatrix} m_1 & -1 & c_1 \\ m_2 & -1 & c_2 \\ m_3 & -1 & c_3 \end{vmatrix} = 0$
Expanding along the second column:
$-(-1) \begin{vmatrix} m_2 & c_2 \\ m_3 & c_3 \end{vmatrix} + (-1) \begin{vmatrix} m_1 & c_1 \\ m_3 & c_3 \end{vmatrix} - (-1) \begin{vmatrix} m_1 & c_1 \\ m_2 & c_2 \end{vmatrix} = 0$
$(m_2c_3 - m_3c_2) - (m_1c_3 - m_3c_1) + (m_1c_2 - m_2c_1) = 0$
Rearranging the terms:
$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

(C) To check if the lines are concurrent,we calculate the determinant of the coefficients:
$D = \begin{vmatrix} p-q & q-r & r-p \\ q-r & r-p & p-q \\ r-p & p-q & q-r \end{vmatrix}$
Applying the operation $C_1 \to C_1 + C_2 + C_3$,the first column becomes:
$(p-q) + (q-r) + (r-p) = 0$
Since the sum of the elements in the first column is $0$,the determinant $D = 0$.
Therefore,the lines are concurrent.

(B) Three lines are concurrent if the determinant of their coefficients is zero.
First,find the point of intersection of $3x + 4y + 6 = 0$ and $6x + 5y + 9 = 0$.
Multiply the first equation by $2$: $6x + 8y + 12 = 0$.
Subtract the second equation from this: $(6x + 8y + 12) - (6x + 5y + 9) = 0 \implies 3y + 3 = 0 \implies y = -1$.
Substitute $y = -1$ into $3x + 4y + 6 = 0$: $3x - 4 + 6 = 0 \implies 3x = -2 \implies x = -2/3$.
The point of intersection is $(-2/3, -1)$.
Now,check which option passes through $(-2/3, -1)$:
For option $(b)$: $3(-2/3) + 3(-1) + 5 = -2 - 3 + 5 = 0$.
Since the point satisfies the equation in option $(b)$,the lines are concurrent.

(A) The lines are concurrent if the determinant of the coefficients is zero:
$\begin{vmatrix} 7 & -8 & 5 \\ 3 & -4 & 5 \\ 4 & 5 & k \end{vmatrix} = 0$
Expanding along the first row:
$7(-4k - 25) - (-8)(3k - 20) + 5(15 - (-16)) = 0$
$7(-4k - 25) + 8(3k - 20) + 5(31) = 0$
$-28k - 175 + 24k - 160 + 155 = 0$
$-4k - 180 = 0$
$-4k = 180$
$k = -45$

(A) The given lines are $L_1: x - 3 = 0$,$L_2: y - 4 = 0$,and $L_3: 4x - 3y + a = 0$.
Since the lines are concurrent,they must intersect at a single point.
First,find the intersection point of $L_1$ and $L_2$ by solving $x = 3$ and $y = 4$.
The point of intersection is $(3, 4)$.
Since the lines are concurrent,this point $(3, 4)$ must satisfy the equation of the third line $L_3: 4x - 3y + a = 0$.
Substituting $x = 3$ and $y = 4$ into the equation:
$4(3) - 3(4) + a = 0$
$12 - 12 + a = 0$
$a = 0$.

(C) Let the lines be $L_1: 15x - 18y + 1 = 0$,$L_2: 12x + 10y - 3 = 0$,and $L_3: 6x + 66y - 11 = 0$.
Three lines are concurrent if there exist constants $a$ and $b$ such that $L_3 = aL_1 + bL_2$.
Consider the linear combination $3(12x + 10y - 3) - 2(15x - 18y + 1) = (36x - 30x) + (30y + 36y) + (-9 - 2) = 6x + 66y - 11$.
This matches $L_3 = 0$.
Since $L_3$ can be expressed as a linear combination of $L_1$ and $L_2$,the lines are concurrent.

(A) The three lines are concurrent if the determinant of their coefficients is zero:
$\left| \begin{array}{ccc} 1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1 \end{array} \right| = 0$
Expanding the determinant along the third row:
$a(2(-5) - 5(-9)) - b(1(-5) - 3(-9)) - 1(1(5) - 3(2)) = 0$
$a(-10 + 45) - b(-5 + 27) - 1(5 - 6) = 0$
$35a - 22b + 1 = 0$
This equation implies that the point $(a, b)$ satisfies the equation of the line $35x - 22y + 1 = 0$.
Therefore,the line passes through the point $(a, b)$.

(B) If the given lines are concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Applying column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\begin{vmatrix} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{vmatrix} = 0$
Expanding the determinant:
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
Dividing the entire equation by $(1-a)(1-b)(1-c)$ (since $a, b, c \neq 1$):
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{-a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$
Adding $1$ to both sides:
$1 - \frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$
$\frac{1-a+a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

(A) The given lines are $ax + 2y + 1 = 0$,$bx + 3y + 1 = 0$,and $cx + 4y + 1 = 0$.
Since these lines are concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$a(3 - 4) - 2(b - c) + 1(4b - 3c) = 0$
$-a - 2b + 2c + 4b - 3c = 0$
$-a + 2b - c = 0$
$2b = a + c$
This condition implies that $a, b, c$ are in $A.P.$

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} 2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(3(-2) - (-3)(2)) - 1(a(-2) - (-3)(3)) - 1(a(2) - 3(3)) = 0$
$2(-6 + 6) - 1(-2a + 9) - 1(2a - 9) = 0$
$2(0) + 2a - 9 - 2a + 9 = 0$
$0 = 0$
Since the determinant is zero regardless of the value of $a$,the lines are concurrent for all values of $a$.

(C) The lines are concurrent if the determinant of their coefficients is zero:
$\left| \begin{array}{ccc} 4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3 \end{array} \right| = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(b)) - 1((1)(5) - (-1)(b)) = 0$
$4(3 - 25) - 3(-3 - 5b) - 1(5 + b) = 0$
$4(-22) + 9 + 15b - 5 - b = 0$
$-88 + 9 + 15b - 5 - b = 0$
$-84 + 14b = 0$
$14b = 84$
$b = 6$

(D) For three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} 3 & -1 & -2 \\ 5 & a & -3 \\ 2 & 1 & -3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$3(-3a - (-3)) - (-1)(-15 - (-6)) + (-2)(5 - 2a) = 0$
$3(-3a + 3) + 1(-9) - 2(5 - 2a) = 0$
$-9a + 9 - 9 - 10 + 4a = 0$
$-5a - 10 = 0$
$-5a = 10$
$a = -2$

(D) For the three lines to be concurrent,the determinant of the coefficients must be zero:
$\begin{vmatrix} l & m & n \\ m & n & l \\ n & l & m \end{vmatrix} = 0$
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\begin{vmatrix} l + m + n & m & n \\ l + m + n & n & l \\ l + m + n & l & m \end{vmatrix} = 0$
Factoring out $(l + m + n)$:
$(l + m + n) \begin{vmatrix} 1 & m & n \\ 1 & n & l \\ 1 & l & m \end{vmatrix} = 0$
Thus,the condition for concurrency is $l + m + n = 0$.

(B) Given the equation of the lines is $4ax + 3by + c = 0$ with the condition $a + b + c = 0$.
From the condition,we can write $c = -(a + b)$.
Substituting this into the line equation:
$4ax + 3by - (a + b) = 0$
Rearranging the terms to group $a$ and $b$:
$a(4x - 1) + b(3y - 1) = 0$
For this equation to hold for all values of $a$ and $b$,the coefficients must be zero:
$4x - 1 = 0 \implies x = \frac{1}{4}$
$3y - 1 = 0 \implies y = \frac{1}{3}$
Thus,the lines are concurrent at the point $(\frac{1}{4}, \frac{1}{3})$.

(C) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$
Substituting the given coefficients:
$\begin{vmatrix} 1 & 0 & q \\ 0 & 1 & -2 \\ 3 & 2 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$1(1 \times 5 - (-2) \times 2) - 0 + q(0 \times 2 - 1 \times 3) = 0$
$1(5 + 4) + q(-3) = 0$
$9 - 3q = 0$
$3q = 9$
$q = 3$

(B) The given lines are $3x + 4y = 5,$ $5x + 4y = 4,$ and $\lambda x + 4y = 6.$
These lines are concurrent if the point of intersection of the first two lines lies on the third line.
Subtracting the first equation from the second: $(5x + 4y) - (3x + 4y) = 4 - 5 \implies 2x = -1 \implies x = -\frac{1}{2}.$
Substituting $x = -\frac{1}{2}$ into $3x + 4y = 5$: $3(-\frac{1}{2}) + 4y = 5 \implies -\frac{3}{2} + 4y = 5 \implies 4y = 5 + \frac{3}{2} = \frac{13}{2} \implies y = \frac{13}{8}.$
The point of intersection is $(-\frac{1}{2}, \frac{13}{8}).$
Since this point lies on $\lambda x + 4y = 6,$ we have $\lambda(-\frac{1}{2}) + 4(\frac{13}{8}) = 6.$
$-\frac{\lambda}{2} + \frac{13}{2} = 6 \implies -\frac{\lambda}{2} = 6 - \frac{13}{2} = -\frac{1}{2} \implies \lambda = 1.$

(A) The given three straight lines are $ax + by - c = 0$,$bx + cy - a = 0$,and $cx + ay - b = 0$.
For these lines to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} a & b & -c \\ b & c & -a \\ c & a & -b \end{vmatrix} = 0$
Taking $-1$ common from the third column:
$-\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$
Applying the operation $C_1 \to C_1 + C_2 + C_3$:
$-(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0$
For this to hold,either $(a+b+c) = 0$ or the determinant $\begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0$.
The condition for concurrency is $a + b + c = 0$.

(D) The equations of the lines are:
$L_1: x + 2y - 3 = 0$
$L_2: 3x + 4y - 7 = 0$
$L_3: 2x + 3y - 4 = 0$
$L_4: 4x + 5y - 6 = 0$
First,check for parallel lines by comparing slopes $(m = -A/B)$:
$m_1 = -1/2, m_2 = -3/4, m_3 = -2/3, m_4 = -4/5$.
Since no slopes are equal,no two lines are parallel. Thus,they cannot form a rectangle.
Next,check for concurrency for any three lines. For $L_1, L_2, L_3$:
$\Delta = \begin{vmatrix} 1 & 2 & -3 \\ 3 & 4 & -7 \\ 2 & 3 & -4 \end{vmatrix} = 1(-16 + 21) - 2(-12 + 14) - 3(9 - 8) = 5 - 4 - 3 = -2 \neq 0$.
Since the determinant is not zero,the lines are not concurrent.
Therefore,the correct option is $D$.

(B) The equation of any line passing through the intersection of $x - y - 4 = 0$ and $3x + y - 7 = 0$ is given by $(x - y - 4) + \lambda(3x + y - 7) = 0$.
This simplifies to $(1 + 3\lambda)x + (\lambda - 1)y - (4 + 7\lambda) = 0$.
Since this line is parallel to $x + 2y = 1$,the ratio of the coefficients of $x$ and $y$ must be equal to the ratio of the coefficients in the given line:
$\frac{1 + 3\lambda}{\lambda - 1} = \frac{1}{2}$.
Cross-multiplying gives $2 + 6\lambda = \lambda - 1$,which implies $5\lambda = -3$,so $\lambda = -\frac{3}{5}$.
Substituting $\lambda = -\frac{3}{5}$ into the family equation:
$(x - y - 4) - \frac{3}{5}(3x + y - 7) = 0$.
Multiplying by $5$: $5x - 5y - 20 - 9x - 3y + 21 = 0$.
This results in $-4x - 8y + 1 = 0$,or $4x + 8y - 1 = 0$.

(C) The given lines are concurrent if and only if the determinant of their coefficients is zero:
$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Expanding the determinant along the first column:
$1(3bc - 4bc) - 1(2ac - 4ac) + 1(2ab - 3ab) = 0$
$-bc + 2ac - ab = 0$
$2ac = ab + bc$
$2ac = b(a + c)$
$b = \frac{2ac}{a + c}$
This is the condition for $a, b,$ and $c$ to be in $H.P.$ (Harmonic Progression).

(B) For the lines to be concurrent,the determinant of the coefficients must be zero:
$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$
Expanding the determinant:
$a(cb - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$
$abc - a^3 - b^3 + abc + abc - c^3 = 0$
$3abc - a^3 - b^3 - c^3 = 0$
$a^3 + b^3 + c^3 - 3abc = 0$
Using the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$
This implies $a + b + c = 0$ or $a^2 + b^2 + c^2 - ab - bc - ca = 0$
Since $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$,the condition is $a + b + c = 0$ (assuming $a, b, c$ are not all equal).

(B) For three lines to be concurrent,the determinant of their coefficients must be zero.
The system of equations is:
$2x + y - 4 = 0$
$3x + 2y - 3 = 0$
$ax + 3y - 2 = 0$
The condition for concurrency is:
$\begin{vmatrix} 2 & 1 & -4 \\ 3 & 2 & -3 \\ a & 3 & -2 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2(-2) - (-3)(3)) - 1(3(-2) - (-3)(a)) - 4(3(3) - 2(a)) = 0$
$2(-4 + 9) - 1(-6 + 3a) - 4(9 - 2a) = 0$
$2(5) + 6 - 3a - 36 + 8a = 0$
$10 + 6 - 36 + 5a = 0$
$-20 + 5a = 0$
$5a = 20$
$a = 4$

(B) The given equation is $ax + by + c = 0$ and the condition is $3a + 2b + 4c = 0$.
From the second equation,we have $c = -\frac{3a + 2b}{4}$.
Substituting this into the first equation:
$ax + by - \frac{3a + 2b}{4} = 0$
Multiply by $4$ to clear the denominator:
$4ax + 4by - 3a - 2b = 0$
Rearranging terms to group $a$ and $b$:
$a(4x - 3) + b(4y - 2) = 0$
For this to be true for all $a$ and $b$,the coefficients must be zero:
$4x - 3 = 0 \implies x = 3/4$
$4y - 2 = 0 \implies y = 2/4 = 1/2$
Thus,the point of concurrency is $(3/4, 1/2)$.

(A) Given the line equation: $ax + by + c = 0$ $(i)$
Given the condition: $3a - 2b + 5c = 0$
Dividing by $5$,we get: $\frac{3}{5}a - \frac{2}{5}b + c = 0$ $(ii)$
Comparing $(i)$ and $(ii)$,we can write the equation as:
$a(x - \frac{3}{5}) + b(y + \frac{2}{5}) = 0$
This represents a family of lines passing through the intersection of $x - \frac{3}{5} = 0$ and $y + \frac{2}{5} = 0$.
Thus,the lines are concurrent at the point $(\frac{3}{5}, -\frac{2}{5})$.
Since the family of lines $L_1 + \lambda L_2 = 0$ is concurrent at the intersection of $L_1$ and $L_2$,$(R)$ is the correct explanation for $(A)$.

(B) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.
Rearranging the terms to isolate $k$:
$2x + kx + y + ky = 5 + 7k$
$(2x + y - 5) + k(x + y - 7) = 0$.
This represents a family of lines passing through the intersection of the lines $2x + y - 5 = 0$ and $x + y - 7 = 0$.
Solving the system of equations:
$2x + y = 5$
$x + y = 7$
Subtracting the second from the first:
$(2x + y) - (x + y) = 5 - 7$
$x = -2$.
Substituting $x = -2$ into $x + y = 7$:
$-2 + y = 7 \implies y = 9$.
Thus,all lines pass through the fixed point $(-2, 9)$.

(A) Let the points be $A(-a, -b)$,$B(0, 0)$,$C(a, b)$,and $D(a^2, ab)$.
To check if points are collinear,we check the slope between them.
Slope of $AB = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a}$.
Slope of $BC = \frac{b - 0}{a - 0} = \frac{b}{a}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B$,and $C$ are collinear.
Now check the slope of $CD = \frac{ab - b}{a^2 - a} = \frac{b(a - 1)}{a(a - 1)} = \frac{b}{a}$ (for $a \neq 1, 0$).
Since the slopes of $AB, BC$,and $CD$ are all equal to $\frac{b}{a}$,all four points $A, B, C$,and $D$ lie on the same straight line.
Therefore,the points are collinear.

(C) The equation of a line passing through the intersection of $L_1: x + y - 3 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $(x + y - 3) + \lambda(2x - y + 1) = 0$.
Since this line passes through the point $(2, -3)$,we substitute $x = 2$ and $y = -3$ into the equation:
$(2 - 3 - 3) + \lambda(2(2) - (-3) + 1) = 0$
$-4 + \lambda(4 + 3 + 1) = 0$
$-4 + 8\lambda = 0$
$8\lambda = 4 \implies \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ back into the family of lines equation:
$(x + y - 3) + \frac{1}{2}(2x - y + 1) = 0$
Multiply by $2$ to clear the fraction:
$2(x + y - 3) + (2x - y + 1) = 0$
$2x + 2y - 6 + 2x - y + 1 = 0$
$4x + y - 5 = 0$.

(C) Step $1$: Find the intersection point of $4x - 3y = 1$ and $5x - 2y = 3$.
Multiplying the first equation by $2$ and the second by $3$: $8x - 6y = 2$ and $15x - 6y = 9$.
Subtracting the equations: $7x = 7$,so $x = 1$.
Substituting $x = 1$ into $4(1) - 3y = 1$,we get $3y = 3$,so $y = 1$.
The intersection point is $(1, 1)$.
Step $2$: The equation of a line parallel to $2x - 3y + 2 = 0$ is of the form $2x - 3y + k = 0$.
Since the line passes through $(1, 1)$,we substitute the coordinates: $2(1) - 3(1) + k = 0$.
$2 - 3 + k = 0$,which gives $k = 1$.
Thus,the required equation is $2x - 3y + 1 = 0$.

(B) The points are collinear if the area of the triangle formed by them is $0$.
$\frac{1}{2} \left| \begin{matrix} 1 & 1 & 1 \\ 0 & \sec^2 \theta & 1 \\ \csc^2 \theta & 0 & 1 \end{matrix} \right| = 0$
Expanding the determinant:
$1(\sec^2 \theta - 0) - 1(0 - \csc^2 \theta) + 1(0 - \sec^2 \theta \csc^2 \theta) = 0$
$\sec^2 \theta + \csc^2 \theta - \sec^2 \theta \csc^2 \theta = 0$
$\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
$\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
$\frac{1}{\sin^2 \theta \cos^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
$0 = 0$
This identity holds for all $\theta$ where $\sec^2 \theta$ and $\csc^2 \theta$ are defined.
Since $\sec^2 \theta$ is undefined at $\theta = \frac{n\pi}{2}$ (where $\cos \theta = 0$) and $\csc^2 \theta$ is undefined at $\theta = n\pi$ (where $\sin \theta = 0$),the points are collinear for all $\theta \neq \frac{n\pi}{2}$.

(A) The family of lines passing through the intersection of $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$ is given by $(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$.
Rearranging the terms,we get $(a + \lambda b)x + (2b - 2a\lambda)y + (3b - 3a\lambda) = 0$.
Since the line is parallel to the $x$-axis,the coefficient of $x$ must be zero.
Therefore,$a + \lambda b = 0$,which gives $\lambda = -a/b$.
Substituting $\lambda = -a/b$ into the equation:
$(ax + 2by + 3b) - \frac{a}{b}(bx - 2ay - 3a) = 0$
$ax + 2by + 3b - ax + \frac{2a^2}{b}y + \frac{3a^2}{b} = 0$
$y(2b + \frac{2a^2}{b}) + (3b + \frac{3a^2}{b}) = 0$
$y(\frac{2b^2 + 2a^2}{b}) = -(\frac{3b^2 + 3a^2}{b})$
$y = -\frac{3(a^2 + b^2)}{2(a^2 + b^2)} = -\frac{3}{2}$.
Thus,the line is $y = -3/2$,which is at a distance of $3/2$ below the $x$-axis.