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Given that:

${}^5C_1 \times \frac{7!}{3!} + {}^5C_2 \times \frac{7!}{2!2!}$

Calculation:

${}^5C_1 \times \frac{7!}{3!} = 5 \times \frac{5040}{6} = 5 \times 840 = 4200$

${}^5C_2 \times \frac{7!}{2!2!} = 10 \times \frac{5040}{4} = 10 \times 1260 = 12600$

⇒ $4200 + 12600 = 16800$

∴ The correct answer is option (3).

Given that:

${}^{36}C_{r+1} = \frac{6({}^{35}C_r)}{k^2 - 3}$

Calculation:

$\frac{36}{r+1} \cdot {}^{35}C_r = 6 \cdot \frac{{}^{35}C_r}{k^2 - 3}$

⇒ $k^2 - 3 = \frac{r+1}{6}$

⇒ $k^2 = \frac{r+19}{6}$

$0 \leq r \leq 35$

⇒ $3.1 \leq \frac{r+19}{6} \leq 9$

⇒ $k^2 = 4, 9$

⇒ $k = \pm 2, \pm 3$

⇒ $(r, k) = (5, \pm 2), (35, \pm 3)$

∴ The correct answer is option (2).

Given that:

$21 = \frac{15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5}{90 + 2k}$

Calculation:

$21(90 + 2k) = 15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5$

⇒ $1890 + 42k = 1910 + 40k$

⇒ $42k - 40k = 1910 - 1890$

⇒ $2k = 20$

⇒ $k = 10$

$k$ satisfies the equation $2x^2 - 23x - 10 = 0$.

∴ The correct answer is option (1).

Given that:

$\text{Midpoints: } \left( \frac{5}{2}, 7 \right), \left( \frac{5}{2}, 3 \right), (4, 5)$

$\text{Incentre: } (h, k)$

Calculation:

$I = \left( \frac{24}{9}, 5 \right)$

⇒ $h = \frac{24}{9} = \frac{8}{3}, ; k = 5$

⇒ $3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$

∴ The correct answer is option (3).

Given that:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,; a < b$

$(4, 3)$ lies on the ellipse

$e = \frac{\sqrt{5}}{3}$

Calculation:

$e^2 = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - e^2 = 1 - \frac{5}{9} = \frac{4}{9}$

$\Rightarrow \frac{a^2}{b^2} = \frac{4}{9} \dots (1)$

$\frac{16}{a^2} + \frac{9}{b^2} = 1 \dots (2)$

From (1) and (2):

$a^2 = 20,; b^2 = 45$

$LR = \frac{2a^2}{b} = \frac{2(20)}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$

∴ The correct answer is option (4).

Given that:

$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$

Calculation:

$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$

⇒ $K = \frac{1}{4} \sin 30^\circ$

⇒ $K = \frac{1}{8}$

$\sin \frac{10K\pi}{3} = \sin\left( 10 \times \frac{1}{8} \times \frac{\pi}{3} \right)$

⇒ $\sin \frac{10K\pi}{3} = \sin \frac{5\pi}{12}$

⇒ $\sin \frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

∴ The correct answer is option (1).

Given that:

$S = { x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbf{Z} }$

Calculation:

$\sin x (\sin x + \cos x) \in \left[ \frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2} \right]$

⇒ $a = 0, 1$ (since $a \in \mathbf{Z}$)

Case 1: $a = 0$

$\sin x (\sin x + \cos x) = 0$

⇒ $\sin x = 0$ or $\sin x + \cos x = 0$

$\sin x = 0 \Rightarrow x = -\pi, 0, \pi$ (3 solutions)

$\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = -\frac{\pi}{4}, \frac{3\pi}{4}$ (2 solutions)

Total solutions for $a = 0$: 3 + 2 = 5

Case 2: $a = 1$

$\sin x (\sin x + \cos x) = 1$

⇒ $\sin^2 x + \sin x \cos x = 1$

⇒ $2\sin^2 x + 2\sin x \cos x = 2$

⇒ $1 - \cos 2x + \sin 2x = 2$

⇒ $\sin 2x - \cos 2x = 1$

Squaring both sides:

$\sin^2 2x - 2\sin 2x \cos 2x + \cos^2 2x = 1$

⇒ $1 - \sin 4x = 1$

⇒ $\sin 4x = 0 \Rightarrow 4x = -4\pi, -3\pi, \dots, 3\pi, 4\pi$

⇒ $x = -\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$

Total solutions for $a = 1$: 4

Total solutions: $5 + 4 = 9$

∴ The correct answer is option (4).

Given that:

$\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$

$\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$

Calculation:

$3r_1 - 1 = r_2 + 2 \Rightarrow r_2 = 3r_1 - 3$

$7r_1 - b - 1 = 0 \Rightarrow b = 7r_1 - 1$

$7r_2 + 2a = 0 \Rightarrow 2a = -7r_2 \Rightarrow a = -\frac{7}{2}r_2$

$r_2 = 3r_1 - 3 \Rightarrow a = -\frac{7}{2}(3r_1 - 3) \Rightarrow a = -\frac{21}{2}r_1 + \frac{21}{2}$

$a + b = -\frac{21}{2}r_1 + \frac{21}{2} + 7r_1 - 1$

$a + b = \left(-\frac{21}{2} + 7\right)r_1 + \left(\frac{21}{2} - 1\right)$

$a + b = \frac{14}{2}r_1 + \frac{19}{2} \Rightarrow r_1 = 1$

$a + b = 7 + 2 = 9$

∴ The correct answer is option (4).

Given that:

$|\vec{a}| = 2, ; |\vec{b}| = 3$

$E = 3\left| \left( 3\vec{a} + 2\vec{b} \right) \right| + 4\left| \left( 3\vec{a} - 2\vec{b} \right) \right|$

Calculation:

$E = 3\sqrt{9a^2 + 4b^2 + 12\vec{a}.\vec{b}} + 4\sqrt{9a^2 + 4b^2 - 12\vec{a}.\vec{b}}$

⇒ $a = 2, ; b = 3, ; \vec{a}.\vec{b} = 6 \cos \theta$

⇒ $E = 3\sqrt{36 + 36 + 12 \times 6 \cos \theta} + 4\sqrt{36 + 36 - 72 \cos \theta}$

⇒ $E = 3\sqrt{72 + 72 \cos \theta} + 4\sqrt{72 - 72 \cos \theta}$

⇒ $E = 18\sqrt{2} \sqrt{1 + \cos \theta} + 24\sqrt{2} \sqrt{1 - \cos \theta}$

⇒ $E = 36 \cos \frac{\theta}{2} + 48 \sin \frac{\theta}{2}$

⇒ $E_{\text{max}} = \sqrt{(36)^2 + (48)^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60$

∴ The correct answer is option (3).

Given that:

$\vec{L}$ passes through point (1, 1, 1) and is perpendicular to $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$.

Calculation:

$\text{Direction vector of } \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$

Equation of line: $\frac{x-1}{-2} = \frac{y-1}{3} = \frac{z-1}{-2} = \lambda$

Let $P(a, b, c) \equiv (2\lambda + 1, -3\lambda + 1, 2\lambda + 1)$ be the foot of perpendicular from (0, 0, 0).

$\vec{OM} \cdot (-2\hat{i} + 3\hat{j} - 2\hat{k}) = 0$

⇒ $-2(2\lambda + 1) + 3(-3\lambda + 1) - 2(2\lambda + 1) = 0$

⇒ $\lambda = -\frac{1}{17}$

Substitute $\lambda = -\frac{1}{17}$:

$a = \frac{-2}{17} + 1, ; b = \frac{3}{17} + 1, ; c = \frac{-2}{17} + 1$

⇒ $34(a+b+c) = 100$

∴ The correct answer is option (3).