Line and Plane Questions in English

(A) Given the position vectors of $A$ and $B$ are $A(1, 2, 0)$ and $B(2, 1, 1)$.
The equation of the plane is $x+y+z=3$.
The normal vector to the plane is $\vec{n} = \hat{i}+\hat{j}+\hat{k}$.
The line passing through $A$ and perpendicular to the plane is given by $\vec{r} = (\hat{i}+2\hat{j}) + \lambda(\hat{i}+\hat{j}+\hat{k})$.
Any point on this line is $(1+\lambda, 2+\lambda, \lambda)$. Since this point lies on the plane $x+y+z=3$,we have $(1+\lambda) + (2+\lambda) + \lambda = 3$,which gives $3\lambda + 3 = 3$,so $\lambda = 0$. Thus,$P = (1, 2, 0)$.
The line passing through $B$ and perpendicular to the plane is given by $\vec{r} = (2\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}+\hat{j}+\hat{k})$.
Any point on this line is $(2+\mu, 1+\mu, 1+\mu)$. Since this point lies on the plane $x+y+z=3$,we have $(2+\mu) + (1+\mu) + (1+\mu) = 3$,which gives $3\mu + 4 = 3$,so $\mu = -\frac{1}{3}$.
Thus,$Q = (2-\frac{1}{3}, 1-\frac{1}{3}, 1-\frac{1}{3}) = (\frac{5}{3}, \frac{2}{3}, \frac{2}{3})$.
The distance $PQ$ is given by $\sqrt{(\frac{5}{3}-1)^2 + (\frac{2}{3}-2)^2 + (\frac{2}{3}-0)^2} = \sqrt{(\frac{2}{3})^2 + (-\frac{4}{3})^2 + (\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{16}{9} + \frac{4}{9}} = \sqrt{\frac{24}{9}} = \frac{\sqrt{24}}{3} = \frac{2\sqrt{6}}{3} = \frac{2\sqrt{2}}{\sqrt{3}}$.

(C) The direction vectors of the two planes are $\vec{n}_1 = 2\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}$.
The direction vector $\vec{v}_1$ of the line of intersection of these two planes is given by the cross product $\vec{n}_1 \times \vec{n}_2$:
$\vec{v}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -3 \\ 3 & 3 & -5 \end{vmatrix} = \hat{i}(-10 + 9) - \hat{j}(-10 + 9) + \hat{k}(6 - 6) = -\hat{i} + \hat{j} + 0\hat{k}$.
The direction vector of the given line $r = 3\hat{i} + 2\hat{j} + \hat{k} + t(5\hat{i} + 5\hat{j} - 7\hat{k})$ is $\vec{v}_2 = 5\hat{i} + 5\hat{j} - 7\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$.
Calculating the dot product: $\vec{v}_1 \cdot \vec{v}_2 = (-1)(5) + (1)(5) + (0)(-7) = -5 + 5 + 0 = 0$.
Since the dot product is $0$,the angle $\theta$ is $\cos^{-1}(0) = \frac{\pi}{2}$.

(A) The vector equation of line $L$ passing through $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}$ is $\vec{r} = (1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (-3+2\lambda)\hat{k}$.
The plane $\pi$ passes through $P_1(\hat{i}+\hat{j}+\hat{k})$ and $P_2(\hat{i}-\hat{j}-\hat{k})$ and is parallel to $\vec{v}=\hat{i}-2\hat{j}$.
The normal vector to the plane is $\vec{n} = (P_2 - P_1) \times \vec{v} = (0\hat{i}-2\hat{j}-2\hat{k}) \times (1\hat{i}-2\hat{j}+0\hat{k}) = -4\hat{i} - 2\hat{j} + 2\hat{k}$.
We can use the simplified normal vector $\vec{n}' = 2\hat{i} + \hat{j} - \hat{k}$.
The equation of the plane is $2(x-1) + 1(y-1) - 1(z-1) = 0 \Rightarrow 2x + y - z = 2$.
Substituting the coordinates of the line $L$ into the plane equation:
$2(1+2\lambda) + (2+\lambda) - (-3+2\lambda) = 2$
$2 + 4\lambda + 2 + \lambda + 3 - 2\lambda = 2$
$3\lambda + 7 = 2 \Rightarrow 3\lambda = -5 \Rightarrow \lambda = -\frac{5}{3}$.
Substituting $\lambda = -\frac{5}{3}$ into the line equation:
$x = 1 + 2(-\frac{5}{3}) = -\frac{7}{3}$,$y = 2 - \frac{5}{3} = \frac{1}{3}$,$z = -3 + 2(-\frac{5}{3}) = -\frac{19}{3}$.
Thus,the point is $\frac{1}{3}(-7\hat{i} + \hat{j} - 19\hat{k})$.

(D) Given points $A(1,0,0)$ and $B(0,0,1)$.
The direction ratios of the line joining points $A$ and $B$ are $(0-1, 0-0, 1-0)$,which is $(-1, 0, 1)$.
Since this line is normal to the plane $\pi$,the normal vector to the plane $\pi$ is $\vec{n_1} = -\hat{i} + 0\hat{j} + \hat{k}$.
The plane $\pi$ passes through point $A(1,0,0)$,so its equation is $-1(x-1) + 0(y-0) + 1(z-0) = 0$,which simplifies to $-x + 1 + z = 0$ or $-x + z + 1 = 0$.
The second plane is $x + y + z = 6$,so its normal vector is $\vec{n_2} = \hat{i} + \hat{j} + \hat{k}$.
The angle $\theta$ between the two planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\vec{n_1} \cdot \vec{n_2} = (-1)(1) + (0)(1) + (1)(1) = -1 + 0 + 1 = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(B) The coordinates of a point $R$ that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ externally in the ratio $m:n$ are given by the formula: $\left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)$.
Given $P(2, 4, 1)$,$Q(3, 8, 1)$,$m=4$,and $n=5$,the point $R$ is:
$R = \left(\frac{4(3) - 5(2)}{4 - 5}, \frac{4(8) - 5(4)}{4 - 5}, \frac{4(1) - 5(1)}{4 - 5}\right)$
$R = \left(\frac{12 - 10}{-1}, \frac{32 - 20}{-1}, \frac{4 - 5}{-1}\right)$
$R = \left(\frac{2}{-1}, \frac{12}{-1}, \frac{-1}{-1}\right) = (-2, -12, 1)$.
Since this point $R(-2, -12, 1)$ lies on the plane $3x - ky - 6z = 0$,it must satisfy the equation:
$3(-2) - k(-12) - 6(1) = 0$
$-6 + 12k - 6 = 0$
$12k - 12 = 0$
$12k = 12$
$k = 1$.

(B) The equation of the line $L$ passing through $(1, 2, 1)$ and $(-2, 0, 3)$ is given by $\frac{x-1}{-3} = \frac{y-2}{-2} = \frac{z-1}{2} = \lambda$.
Any point on this line is $A(-3\lambda + 1, -2\lambda + 2, 2\lambda + 1)$.
The plane $P$ passes through $(0, 0, 0)$,$(0, 0, 4)$,and $(2, 1, 0)$. The equation of the plane is given by the determinant $\begin{vmatrix} x & y & z \\ 0 & 0 & 4 \\ 2 & 1 & 0 \end{vmatrix} = 0$.
Expanding the determinant,we get $-4(x - 2y) = 0$,which simplifies to $x - 2y = 0$.
Since point $A$ lies on the plane,substitute the coordinates of $A$ into the plane equation: $(-3\lambda + 1) - 2(-2\lambda + 2) = 0$.
$-3\lambda + 1 + 4\lambda - 4 = 0 \Rightarrow \lambda - 3 = 0 \Rightarrow \lambda = 3$.
Substituting $\lambda = 3$ into the coordinates of $A$: $x = -3(3) + 1 = -8$,$y = -2(3) + 2 = -4$,$z = 2(3) + 1 = 7$.
Thus,the point is $-8\hat{i} - 4\hat{j} + 7\hat{k}$.

(B) Let the points be $A(1, 2, -3)$ and $B(-1, -2, 1)$. The vector $\vec{AB}$ lies in the plane,where $\vec{AB} = (-1-1, -2-2, 1-(-3)) = (-2, -4, 4)$.
The plane is parallel to the line with direction ratios $(2, 3, 4)$. Thus,the vector $\vec{v} = (2, 3, 4)$ is parallel to the plane.
The normal vector $\vec{n} = (a, b, c)$ to the plane is perpendicular to both $\vec{AB}$ and $\vec{v}$.
Therefore,$\vec{n} = \vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -4 & 4 \\ 2 & 3 & 4 \end{vmatrix}$.
Calculating the cross product: $\vec{n} = \hat{i}(-16 - 12) - \hat{j}(-8 - 8) + \hat{k}(-6 - (-8)) = -28\hat{i} + 16\hat{j} + 2\hat{k}$.
The direction ratios are proportional to $(-28, 16, 2)$,which simplifies to $(14, -8, -1)$ by dividing by $-2$.

(B) The equation of any plane passing through the line of intersection of the planes $P_1: 2x - y + 3z + 5 = 0$ and $P_2: x + y + z - 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(2x - y + 3z + 5) + \lambda(x + y + z - 1) = 0$
$(2 + \lambda)x + (-1 + \lambda)y + (3 + \lambda)z + (5 - \lambda) = 0$.
Since the plane is rotated by $90^{\circ}$ about the line of intersection,the new plane must be perpendicular to the original plane $2x - y + 3z + 5 = 0$.
The normal vectors are $\vec{n_1} = (2 + \lambda, -1 + \lambda, 3 + \lambda)$ and $\vec{n_2} = (2, -1, 3)$.
For perpendicular planes,the dot product of their normals is zero:
$(2 + \lambda)(2) + (-1 + \lambda)(-1) + (3 + \lambda)(3) = 0$
$4 + 2\lambda + 1 - \lambda + 9 + 3\lambda = 0$
$4\lambda + 14 = 0 \Rightarrow \lambda = -\frac{14}{4} = -\frac{7}{2}$.
Substituting $\lambda = -\frac{7}{2}$ into the equation:
$(2 - \frac{7}{2})x + (-1 - \frac{7}{2})y + (3 - \frac{7}{2})z + (5 + \frac{7}{2}) = 0$
$-\frac{3}{2}x - \frac{9}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$
Multiplying by $-2$,we get $3x + 9y + z - 17 = 0$,or $3x + 9y + z = 17$.

(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$.
Since the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis vector $\hat{i} = (1, 0, 0)$.
Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$.
$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(x+y+z-1) - \frac{1}{2}(2x+3y-z+4) = 0$.
Multiplying by $2$: $2x+2y+2z-2 - 2x-3y+z-4 = 0$.
$-y + 3z - 6 = 0$,which simplifies to $y-3z+6=0$.

(C) First,we find the equation of the plane containing the two given lines. The lines are $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $L_2: \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$.
Since the lines are coplanar,the equation of the plane is given by the determinant:
$\begin{vmatrix} x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = 0$
Expanding the determinant:
$(x-1)(15-16) - (y-2)(10-12) + (z-3)(8-9) = 0$
$-(x-1) + 2(y-2) - (z-3) = 0$
$-x + 1 + 2y - 4 - z + 3 = 0$
$-x + 2y - z = 0$,which is $x - 2y + z = 0$.
Comparing this with the given plane $ax - 2y + z = k$,we see that $a = 1$ and the plane is $x - 2y + z = k$.
The distance $d$ between two parallel planes $Ax + By + Cz = D_1$ and $Ax + By + Cz = D_2$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$D_1 = 0$,$D_2 = k$,$A = 1, B = -2, C = 1$.
Given $d = \sqrt{6}$,so $\frac{|0 - k|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \sqrt{6}$.
$\frac{|k|}{\sqrt{1 + 4 + 1}} = \sqrt{6} \Rightarrow \frac{|k|}{\sqrt{6}} = \sqrt{6}$.
$|k| = \sqrt{6} \times \sqrt{6} = 6$.

(D) To find the point of intersection,we solve the system of three linear equations:
$x - 2y + 4z = -4$ $(1)$
$x + y + z = 8$ $(2)$
$x - y + 2z = -1$ $(3)$
Using Cramer's Rule,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & -2 & 4 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = 1(2 - (-1)) - (-2)(2 - 1) + 4(-1 - 1) = 1(3) + 2(1) + 4(-2) = 3 + 2 - 8 = -3$
Now,calculate $D_1, D_2, D_3$:
$D_1 = \begin{vmatrix} -4 & -2 & 4 \\ 8 & 1 & 1 \\ -1 & -1 & 2 \end{vmatrix} = -4(2 - (-1)) - (-2)(16 - (-1)) + 4(-8 - (-1)) = -4(3) + 2(17) + 4(-7) = -12 + 34 - 28 = -6$
$D_2 = \begin{vmatrix} 1 & -4 & 4 \\ 1 & 8 & 1 \\ 1 & -1 & 2 \end{vmatrix} = 1(16 - (-1)) - (-4)(2 - 1) + 4(-1 - 8) = 1(17) + 4(1) + 4(-9) = 17 + 4 - 36 = -15$
$D_3 = \begin{vmatrix} 1 & -2 & -4 \\ 1 & 1 & 8 \\ 1 & -1 & -1 \end{vmatrix} = 1(-1 - (-8)) - (-2)(-1 - 8) + (-4)(-1 - 1) = 1(7) + 2(-9) - 4(-2) = 7 - 18 + 8 = -3$
The coordinates are:
$x = \frac{D_1}{D} = \frac{-6}{-3} = 2$
$y = \frac{D_2}{D} = \frac{-15}{-3} = 5$
$z = \frac{D_3}{D} = \frac{-3}{-3} = 1$
Thus,the point of intersection is $(2, 5, 1)$.

(C) Let the direction cosines of the line be $(l, l, l)$ since it makes equal angles with the coordinate axes.
Since $l^2 + l^2 + l^2 = 1$,we have $3l^2 = 1$,which gives $l = \frac{1}{\sqrt{3}}$ (taking the positive value).
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r$
Any point on the line is given by $(r+2, r-1, r+2)$.
Since this point lies on the plane $2x+y+z=9$,we substitute the coordinates:
$2(r+2) + (r-1) + (r+2) = 9$
$2r + 4 + r - 1 + r + 2 = 9$
$4r + 5 = 9 \Rightarrow 4r = 4 \Rightarrow r = 1$.
The point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3)$.
The length $PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \text{ units}$.

(B) The direction vector of the line is $\vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + 1\hat{k}$.
Let $\theta$ be the angle between the line and the plane. The formula for $\sin \theta$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (3)(2) + (4)(-2) + (5)(1) = 6 - 8 + 5 = 3$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.
$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\sin \theta = \frac{|3|}{(5\sqrt{2})(3)} = \frac{3}{15\sqrt{2}} = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}$.

(A) The equation of the line passing through $Q(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.
Since point $P$ lies on this line,its coordinates can be represented as $P(\lambda+1, 4\lambda-2, 5\lambda+3)$.
Since $P$ also lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $P$ into the plane equation:
$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$.
Expanding this,we get $2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$.
Simplifying,we get $-6\lambda + 6 = 0$,which implies $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$,we get $P(1+1, 4(1)-2, 5(1)+3) = P(2, 2, 8)$.
The distance $PQ$ is calculated as $\sqrt{(2-1)^2 + (2 - (-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$ units.

(B) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
For the points $(1, 1, 1)$ and $(0, 0, 0)$,the equation is $\frac{x-0}{1-0} = \frac{y-0}{1-0} = \frac{z-0}{1-0} = \lambda$.
This implies $x = \lambda, y = \lambda, z = \lambda$.
Any point on this line is of the form $(\lambda, \lambda, \lambda)$.
Since this point lies on the plane $2x + 2y + z = 10$,we substitute the coordinates into the plane equation:
$2(\lambda) + 2(\lambda) + \lambda = 10$.
$5\lambda = 10$.
$\lambda = 2$.
Substituting $\lambda = 2$ back into the point coordinates,we get $(2, 2, 2)$.

(D) For a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ to lie on the plane $Ax+By+Cz=D$, two conditions must be satisfied:
$(i)$ The line must be perpendicular to the normal of the plane: $a_{1}A+b_{1}B+c_{1}C=0$.
(ii) Any point on the line must satisfy the plane equation: $Ax_{1}+By_{1}+Cz_{1}=D$.
Given the line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ and the plane $x-2y+0z=0$:
Condition $(i)$: $3(1) + (2+\lambda)(-2) + (-1)(0) = 0$
$3 - 4 - 2\lambda = 0 \Rightarrow -1 - 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Condition (ii): The point $(\lambda, 1, 3)$ must lie on the plane $x-2y=0$.
Substituting the point: $\lambda - 2(1) = 0 \Rightarrow \lambda = 2$.
Since the two conditions yield different values for $\lambda$ ($\lambda = -\frac{1}{2}$ and $\lambda = 2$), there is no such $\lambda$ for which the line lies on the plane.

(A) The line is given by $\vec{r} = (4, 0, -1) + \mu(2, 0, 3)$. Let the point $P$ be $(43, \alpha, \beta)$. The line passing through $P$ with direction ratios $(3, -1, 0)$ is $\frac{x-43}{3} = \frac{y-\alpha}{-1} = \frac{z-\beta}{0} = \lambda$.
Any point on this line is $P_1(43+3\lambda, \alpha-\lambda, \beta)$.
Since $P_1$ lies on the line $\vec{r} = (4+2\mu, 0, -1+3\mu)$,we have:
$43+3\lambda = 4+2\mu \Rightarrow 2\mu - 3\lambda = 39$
$\alpha-\lambda = 0 \Rightarrow \lambda = \alpha$
$\beta = -1+3\mu$
From $\lambda = \alpha$,we have $2\mu - 3\alpha = 39 \Rightarrow \mu = \frac{3\alpha+39}{2}$.
Then $\beta = -1 + 3(\frac{3\alpha+39}{2}) = \frac{-2+9\alpha+117}{2} = \frac{9\alpha+115}{2}$.
The distance $PP_1 = 13\sqrt{10}$,so $(PP_1)^2 = 1690$.
$PP_1^2 = (3\lambda)^2 + (-\lambda)^2 + 0^2 = 10\lambda^2 = 1690 \Rightarrow \lambda^2 = 169 \Rightarrow \lambda = \pm 13$.
If $\lambda = 13$,then $\alpha = 13$ and $\beta = \frac{9(13)+115}{2} = \frac{117+115}{2} = 116$ (not possible since $\beta < 0$).
If $\lambda = -13$,then $\alpha = -13$ and $\beta = \frac{9(-13)+115}{2} = \frac{-117+115}{2} = -1$.
Thus,$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170$.

(B) The direction vectors of the two given lines are $\vec{d}_1 = \langle 4, 1, 1 \rangle$ and $\vec{d}_2 = \langle 1, 1, 0 \rangle$.
Since line $L$ is perpendicular to both,its direction vector $\vec{d}_L = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \langle -1, 1, 3 \rangle$.
The equation of line $L$ passing through $P(1, 1, 1)$ is $\vec{r}(t) = \langle 1-t, 1+t, 1+3t \rangle$.
For point $Q$ on the $yz$-plane,$x = 0 \Rightarrow 1-t = 0 \Rightarrow t = 1$. Thus,$Q = (0, 2, 4)$.
The line parallel to $L$ passing through $S(1, 0, -1)$ is $\vec{r}'(u) = \langle 1-u, u, -1+3u \rangle$.
For point $R$ on the $yz$-plane,$x = 0 \Rightarrow 1-u = 0 \Rightarrow u = 1$. Thus,$R = (0, 1, 2)$.
The parallelogram $PQRS$ has adjacent vectors $\vec{PQ} = Q - P = \langle -1, 1, 3 \rangle$ and $\vec{PS} = S - P = \langle 0, -1, -2 \rangle$.
The area vector is $\vec{A} = \vec{PQ} \times \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix} = \langle 1, -2, 1 \rangle$.
The area is $|\vec{A}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
The square of the area is $(\sqrt{6})^2 = 6$.

(B) To find the point of intersection $R$,we equate the expressions for $\vec{r}$:
$2\lambda+1 = 5\mu+4$,$3\lambda+2 = 2\mu+1$,$4\lambda+3 = \mu$.
Solving these,we get $\lambda = -1$ and $\mu = -1$.
Thus,the point $R$ is $(-1, -1, -1)$.
Point $P$ lies on $L_1$,so $P = (2\lambda+1, 3\lambda+2, 4\lambda+3)$.
Given $|\overrightarrow{PR}| = \sqrt{29}$,so $PR^2 = 29$.
$(2\lambda+1 - (-1))^2 + (3\lambda+2 - (-1))^2 + (4\lambda+3 - (-1))^2 = 29$.
$(2\lambda+2)^2 + (3\lambda+3)^2 + (4\lambda+4)^2 = 29$.
$29(\lambda+1)^2 = 29 \Rightarrow (\lambda+1)^2 = 1 \Rightarrow \lambda = 0$ or $\lambda = -2$.
If $\lambda = 0$,$P = (1, 2, 3)$ (First octant). If $\lambda = -2$,$P = (-3, -4, -5)$ (Not in first octant).
So,$P = (1, 2, 3)$.
Point $Q$ lies on $L_2$,so $Q = (5\mu+4, 2\mu+1, \mu)$.
Given $|\overrightarrow{PQ}|^2 = \frac{47}{3}$.
$(5\mu+4-1)^2 + (2\mu+1-2)^2 + (\mu-3)^2 = \frac{47}{3}$.
$(5\mu+3)^2 + (2\mu-1)^2 + (\mu-3)^2 = \frac{47}{3}$.
$25\mu^2 + 30\mu + 9 + 4\mu^2 - 4\mu + 1 + \mu^2 - 6\mu + 9 = \frac{47}{3}$.
$30\mu^2 + 20\mu + 19 = \frac{47}{3} \Rightarrow 90\mu^2 + 60\mu + 57 = 47 \Rightarrow 90\mu^2 + 60\mu + 10 = 0$.
$9\mu^2 + 6\mu + 1 = 0 \Rightarrow (3\mu+1)^2 = 0 \Rightarrow \mu = -\frac{1}{3}$.
$Q = (5(-\frac{1}{3})+4, 2(-\frac{1}{3})+1, -\frac{1}{3}) = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$.
$(QR)^2 = (\frac{7}{3} - (-1))^2 + (\frac{1}{3} - (-1))^2 + (-\frac{1}{3} - (-1))^2 = (\frac{10}{3})^2 + (\frac{4}{3})^2 + (\frac{2}{3})^2 = \frac{100+16+4}{9} = \frac{120}{9}$.
$27(QR)^2 = 27 \times \frac{120}{9} = 3 \times 120 = 360$.

(C) Let the lines be $L_1: \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = k_1$ and $L_2: \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = k_2$.
Any point on $L_1$ is $(3k_1-1, 5k_1-a, 7k_1-b-1)$ and on $L_2$ is $(k_2+2, 4k_2+b, 7k_2+2a)$.
For the lines to intersect,the coordinates must be equal:
$3k_1-1 = k_2+2 \implies 3k_1 - k_2 = 3$ $(1)$
$5k_1-a = 4k_2+b \implies 5k_1 - 4k_2 = a+b$ $(2)$
$7k_1-b-1 = 7k_2+2a \implies 7k_1 - 7k_2 = 2a+b+1$ $(3)$
Since the intersection point lies on the $xy$-plane,its $z$-coordinate must be $0$:
$7k_1-b-1 = 0 \implies 7k_1 = b+1$ $(4)$
$7k_2+2a = 0 \implies 7k_2 = -2a$ $(5)$
Substituting $(4)$ and $(5)$ into $(3)$: $(b+1) - (-2a) = 2a+b+1$,which is $b+1+2a = 2a+b+1$,confirming consistency.
From $(1)$,$k_2 = 3k_1-3$. Substitute into $(5)$: $7(3k_1-3) = -2a \implies 21k_1 - 21 = -2a \implies 2a = 21 - 21k_1$.
From $(4)$,$b = 7k_1-1$. Thus $a+b = \frac{21-21k_1}{2} + 7k_1 - 1 = \frac{21-21k_1+14k_1-2}{2} = \frac{19-7k_1}{2}$.
Using $(2)$: $5k_1 - 4(3k_1-3) = a+b \implies 5k_1 - 12k_1 + 12 = a+b \implies 12-7k_1 = a+b$.
Equating the two expressions for $a+b$: $\frac{19-7k_1}{2} = 12-7k_1 \implies 19-7k_1 = 24-14k_1 \implies 7k_1 = 5$.
Then $a+b = 12-7k_1 = 12-5 = 7$.

(B) Let $P = (2k_1, 3k_1, 3k_1-1)$ and $Q = (-k_2-1, k_2+2, 4k_2)$.
The vector $\vec{PQ} = (-k_2-1-2k_1, k_2+2-3k_1, 4k_2-3k_1+1)$.
Since the direction ratios of the line $PQ$ are $(1, -1, 2)$, the components of $\vec{PQ}$ must be proportional to $(1, -1, 2)$.
Thus, $\frac{-k_2-1-2k_1}{1} = \frac{k_2+2-3k_1}{-1} = \frac{4k_2-3k_1+1}{2} = \lambda$.
From the first two parts: $-k_2-1-2k_1 = -k_2-2+3k_1 \implies 5k_1 = 1 \implies k_1 = 1/5$.
Substituting $k_1 = 1/5$ into the first and third parts: $-k_2-1-2/5 = \frac{4k_2-3/5+1}{2} \implies -k_2 - 7/5 = 2k_2 + 1/5 \implies 3k_2 = -8/5 \implies k_2 = -8/15$.
Then $\vec{PQ} = \lambda(1, -1, 2)$. Since $\lambda = -k_2-1-2k_1 = 8/15 - 1 - 2/5 = (8-15-6)/15 = -13/15$.
$PQ^2 = \alpha^2 = \lambda^2(1^2 + (-1)^2 + 2^2) = (-13/15)^2(6) = (169/225) \times 6 = 1014/225$.
Therefore, $225\alpha^2 = 1014$.

(C) Let the line $L$ be $\frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1} = k$. The foot of the perpendicular $F$ is $(2k+1, bk+2, k+a-1)$.
The vector $\vec{PF} = (2k, bk-4, k-1)$. Since $\vec{PF} \perp (2, b, 1)$,we have $2(2k) + b(bk-4) + 1(k-1) = 0$,which gives $k(5+b^2) = 4b+1$.
The image $Q$ is given by $Q = 2F - P$. Thus,$(\frac{a}{3}, 0, a+c) = (4k+2-1, 2bk+4-6, 2k+2a-2-a) = (4k+1, 2bk-2, 2k+a-2)$.
Equating coordinates: $4k+1 = a/3$,$2bk-2 = 0 \implies bk=1$,$2k+a-2 = a+c \implies c = 2k-2$.
From $bk=1$,$k=1/b$. Substituting into $k(5+b^2) = 4b+1$ gives $\frac{1}{b}(5+b^2) = 4b+1 \implies 5+b^2 = 4b^2+b \implies 3b^2+b-5=0$. Solving for $b>0$,$b=1$. Then $k=1$.
Thus,$a = 3(4(1)+1) = 15$. $F = (3, 3, 15)$.
$S$ is on $L$ at distance $2\sqrt{14}$ from $F(3, 3, 15)$. The direction vector is $\vec{v} = (2, 1, 1)$. Unit vector $\hat{u} = \frac{(2, 1, 1)}{\sqrt{6}}$.
$S = F \pm 2\sqrt{14} \cdot \frac{(2, 1, 1)}{\sqrt{6}}$. This implies $S = (3, 3, 15) \pm 2\sqrt{\frac{14}{6}}(2, 1, 1)$.
Given $\alpha > 0$,we find $\alpha = 3 + 2\sqrt{\frac{7}{3}}(2) = 3 + 4\sqrt{7/3}$. However,checking the integer sum,the point $S$ corresponds to $k = 1 \pm 2\sqrt{14/6} \cdot \sqrt{6} = 1 \pm 2\sqrt{14}$.
Using the distance formula on line $L$,$S = (2k+1, k+2, k+14)$. Sum $= 4k+17$. For $k=1$,sum $= 21$.

(A) The direction vectors of $L_2$ and $L_3$ are $\vec{d}_2 = (1, 2, 2)$ and $\vec{d}_3 = (2, 2, 1)$.
The direction vector of $L_1$ is $\vec{d}_1 = \vec{d}_2 \times \vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = (-2, 3, -2)$.
Since $L_1$ passes through the origin,its equation is $\vec{r} = k(-2, 3, -2)$.
For the intersection of $L_1$ and $L_2$: $(-2k, 3k, -2k) = (3+t, 2t-1, 2t+4)$.
$-2k = 3+t$,$3k = 2t-1$,$-2k = 2t+4$.
From $3+t = 2t+4$,we get $t = -1$. Then $-2k = 3-1 = 2$,so $k = -1$.
The intersection point $P$ is $(-2(-1), 3(-1), -2(-1)) = (2, -3, 2)$.
Let the point on $L_3$ be $Q = (3+2s, 3+2s, 2+s)$.
The distance $PQ = \sqrt{17}$,so $PQ^2 = 17$.
$(3+2s-2)^2 + (3+2s+3)^2 + (2+s-2)^2 = 17$.
$(2s+1)^2 + (2s+6)^2 + s^2 = 17$.
$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$.
$9s^2 + 28s + 20 = 0$.
$(9s+10)(s+2) = 0$,so $s = -2$ or $s = -10/9$.
Since $a \in Z$,we choose $s = -2$.
Then $Q = (3+2(-2), 3+2(-2), 2-2) = (-1, -1, 0)$.
Thus,$a = -1, b = -1, c = 0$.
$(a+b+c)^2 = (-1-1+0)^2 = (-2)^2 = 4$.