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$\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}$

Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given)

$\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{PA} \mathrm{P}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$

Now post multiply by P

$\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}$

$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$

Similarly $A^{10}=P^T B^{10} P=C$

$\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text { (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}$

Similarly check $\mathrm{A}^3$ and so on since $\mathrm{C}=\mathrm{A}^{10}$

$\Rightarrow$ Sum of diagonal elements of C is $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$

$\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{~g} \mathrm{~cd}(\mathrm{~m}, \mathrm{n})=1 \text { (Given) } \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{M}_{A B}=0 \Rightarrow \mathrm{OM} \text { is vertical } \\ & \Rightarrow \alpha=2 \\ & \therefore \text { Centre }(0) \equiv(2,-4) \\ & \quad r=O A=\sqrt{(2-4)^2+(2+4)^2}=\sqrt{40} \end{aligned}\\ &\text { mid point of chord is } \mathrm{N} \equiv(1,2) \quad \therefore \mathrm{ON}=\sqrt{37}\\ &\begin{aligned} \therefore \text { length of chord } & =2 \sqrt{\mathrm{r}^2-(\mathrm{ON})^2} \\ & =2 \sqrt{40-37}=2 \sqrt{3} \end{aligned} \end{aligned}$

Step 1. Identify the Given Lines

The line

$ \mathrm{L}_1:\; \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2} $

passes through the point

$ P_1=(1,2,1) $

with direction vector

$ \mathbf{u}=(1,-1,2). $

Similarly, the line

$ \mathrm{L}_2:\; \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} $

passes through the point

$ P_2=(-1,2,0) $

with direction vector

$ \mathbf{v}=(-1,2,1). $

Step 2. Determine the Direction of $ L_3 $

Since $ L_3 $ is perpendicular to both $ L_1 $ and $ L_2 $, its direction vector must be parallel to the cross product of $ \mathbf{u} $ and $ \mathbf{v} $. Compute:

$ \mathbf{d} = \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}. $

Using the determinant, we obtain:

$ \begin{aligned} d_x &= (-1)(1) - (2)(2) = -1-4=-5,\\[1mm] d_y &= (2)(-1) - (1)(1) = -2-1=-3,\\[1mm] d_z &= (1)(2) - (-1)(-1) = 2-1=1. \end{aligned} $

Thus, a valid direction vector is

$ \mathbf{d}=(-5,-3,1). $

Step 3. Express $ L_3 $ in Terms of Its Intersection with $ L_1 $

Since $ L_3 $ intersects $ L_1 $, let the intersection point (on $ L_1 $) be expressed using a parameter $ t $ as:

$ A(t) = (1+t,\, 2-t,\, 1+2t). $

Since $ L_3 $ passes through $ A(t) $ and also through the given point

$ Q=(\alpha,\beta,\gamma), $

its parametric form can be written as:

$ (\alpha,\beta,\gamma)=A(t) + s\,\mathbf{d} = (1+t,\,2-t,\,1+2t) + s(-5,-3,1) $

for some parameters $ s $ and $ t $.

Step 4. Express $\alpha,\beta,\gamma$ in Terms of $t$ and $s$

Comparing coordinates, we have:

$ \begin{cases} \alpha = 1 + t - 5s,\\[1mm] \beta = 2 - t - 3s,\\[1mm] \gamma = 1 + 2t + s. \end{cases} $

Step 5. Compute $5\alpha - 11\beta - 8\gamma$

Substitute the expressions for $\alpha$, $\beta$, and $\gamma$:

$ \begin{aligned} 5\alpha - 11\beta - 8\gamma &= 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)\\[1mm] &= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)\\[1mm] &= \left(5 - 22 - 8\right) + \left(5t + 11t - 16t\right) + \left(-25s + 33s - 8s\right)\\[1mm] &= -25 + 0t + 0s\\[1mm] &= -25. \end{aligned} $

Taking the absolute value yields:

$ \left|5\alpha-11\beta-8\gamma\right| = 25. $

Conclusion

The value of

$ \left|5\,\alpha - 11\,\beta - 8\,\gamma\right| $ is $ \boxed{25}. $

We begin with the system:

$ \begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned} $

Step 1. Solve the first equation for $x$:

$ x = 6 - y - 2z. $

Step 2. Substitute $x = 6-y-2z$ into the second equation:

$ 2(6-y-2z) + 3y + az = a+1. $

Expanding and simplifying:

$ 12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11. $

Call this Equation (I).

Step 3. Substitute $x = 6-y-2z$ into the third equation:

$ -(6-y-2z) - 3y + bz = 2b. $

Expanding and simplifying:

$ -6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6. $

Call this Equation (II).

Step 4. For the system to have infinitely many solutions, the two equations in $y$ and $z$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $k$ such that

$ -2 = k \cdot 1 \quad \Longrightarrow \quad k = -2. $

Apply this to the coefficient of $z$ and the constant term.

For the $z$-coefficient in Equations (I) and (II):

$ b+2 = k(a-4) = -2(a-4) = -2a+8. $

Thus,

$ b = -2a+6. $

For the constant term:

$ 2b+6 = k(a-11) = -2(a-11) = -2a + 22. $

Substitute $b = -2a+6$ into this equation:

$ 2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22. $

Simplify:

$ -4a + 18 = -2a + 22. $

Solve for $a$:

$ -4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22, $

$ 2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2. $

Substitute $a = -2$ into $b = -2a+6$:

$ b = -2(-2) + 6 = 4 + 6 = 10. $

Step 5. We now compute

$ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. $

Thus, the value of $7a+3b$ is

$ \boxed{16}. $

$\begin{aligned} & \text { General term }={ }^n C_r\left(7^{1 / 3}\right)^{n-r}\left(11^{1 / 12}\right)^r \\ & ={ }^n C_r(7)^{\frac{n-r}{3}}(11)^{r / 12} \end{aligned}$

For integral terms, $r$ must be multiple of 12

$\therefore \mathrm{r}=12 \mathrm{k}, \mathrm{k} \in \mathrm{~W}$

Total values of $\mathrm{r}=183$

Hence $\max r=12(182)$

$=2184$

Min value of $n=2184$