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Calculation:

We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$ ... (i)

And $\cos(\alpha + \beta) \cdot \cos(\alpha - \beta) = \cos^{2} \alpha - \sin^{2} \beta$

Substitute and simplify:

$= \cos^{2} \alpha + \cos^{2} \beta - 1 = 1 - \cos^{2} \gamma \quad \Rightarrow \quad - \cos^{2} \gamma$

Hence, the correct answer is Option 2.

Calculation:

Given,

Points A(1, 2, -1), B(2, 5, -2), and C(4, 4, -3) are three vertices of the rectangle.

We need to find the area of the rectangle formed by the vectors $AB$ and $BC$.

Length of vector $AB$:

$\text{Length of AB} = \sqrt{(2 - 1)^2 + (5 - 2)^2 + (-2 - (-1))^2}$

$= \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11}$

Length of vector $BC$:

$\text{Length of BC} = \sqrt{(4 - 2)^2 + (4 - 5)^2 + (-3 - (-2))^2}$

$= \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}$

Area of the rectangle is the product of the lengths of vectors $AB$ and $BC$:

$\text{Area} = \sqrt{11} \times \sqrt{6} = \sqrt{66}$

∴ The area of the rectangle is $\sqrt{66}$ square units.

Hence, the correct answer is Option 3.

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

$\overrightarrow{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3)$

$\overrightarrow{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1)$

The dot product of $\overrightarrow{AB}$ and $\overrightarrow{BC}$ is:

$\overrightarrow{AB} \cdot \overrightarrow{BC} = (k)(2) + (-1)(k) + (3)(-1) = 2k - k - 3 = k - 3$

Since the vectors are perpendicular, their dot product must be zero:

$k - 3 = 0$

Thus, $k = 3$.

∴ The value of k is $3$.

Hence, the correct answer is Option 4.

Calculation:

Given,

The line equation is: $\frac{x+1}{p} = \frac{y-1}{q} = \frac{z-2}{r}$,  where p = 2q = 3r .

The line makes an angle θ  with the positive direction of the y-axis.

Using the relation between the direction cosines

$l = \frac{p}{\sqrt{p^2 + q^2 + r^2}}, m = \frac{q}{\sqrt{p^2 + q^2 + r^2}}, n = \frac{r}{\sqrt{p^2 + q^2 + r^2}}$,

we substitute p = 3r  and  q = $\frac{3r}{2}$

Direction cosine with y -axis is given by $m = \frac{q}{\sqrt{p^2 + q^2 + r^2}}$.

$m = \frac{\frac{3r}{2}}{\sqrt{(3r)^2 + \left(\frac{3r}{2}\right)^2 + r^2}} = \frac{\frac{3r}{2}}{\sqrt{9r^2 + \frac{9r^2}{4} + r^2}} = \frac{\frac{3r}{2}}{\sqrt{\frac{49r^2}{4}}} = \frac{3}{7}$.

Now, using the double angle identity for cosine $\cos 2θ = 2 \cos^2 θ - 1$:

$\cos 2θ = 2 \left( \frac{3}{7} \right)^2 - 1 = 2 \times \frac{9}{49} - 1 = \frac{18}{49} - 1 = \frac{-31}{49}$.

∴ The value of $\cos 2θ$ is $-\frac{31}{49}$.

Hence, the correct answer is Option 1.

Calculation:

Given,

The plane passes through the point $P(1, 1, 1)$ and is perpendicular to the line with direction ratios $(3, 2, 1)$.

The general equation of the plane is:

$Ax + By + Cz = D$, where $(A, B, C)$ are the direction ratios of the normal to the plane.

Since the plane is perpendicular to the line, the normal vector to the plane is $(3, 2, 1)$.

Thus, the equation of the plane becomes:

$3x + 2y + z = D$.

To find $D$, substitute the point $(1, 1, 1)$ into the equation of the plane:

$3(1) + 2(1) + 1 = D$, which simplifies to:

$3 + 2 + 1 = D$,

$D = 6$.

∴ The equation of the plane is $3x + 2y + z = 6$.

Hence, the correct answer is Option 2.

Calculation:

Given,

$\cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1$

Using the identity $\cos^2(x) = 1 - \sin^2(x)$, we substitute:

$(1 - \sin^2(\alpha)) + (1 - \sin^2(\beta)) + (1 - \sin^2(\gamma)) = 1$

Simplifying the equation:

$3 - (\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma)) = 1$

Rearrange to isolate the sine terms:

$\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2$

Now, calculate the dot product:

$\vec{a} \cdot \vec{b} = \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2$

∴ The value of $\vec{a} \cdot \vec{b}$is 2.

Hence, the correct answer is Option 4.

Calculation:

Given,

The vector $\vec{d} = (\vec{a} \times \vec{b}) \times \vec{c}$

Statement I: $\vec{d}$ is coplanar with $\vec{a}$ and $\vec{b}$.

We use the vector triple product identity: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$.

This shows that $\vec{d}$ is a linear combination of $\vec{a}$ and $\vec{b}$, hence $\vec{d}$ is coplanar with $\vec{a}$ and $\vec{b}$.

Therefore, Statement I is correct.

Statement II: $\vec{d}$ is perpendicular to $\vec{c}$.

To check this, compute the dot product $\vec{d} \cdot \vec{c}$. Using the vector triple product identity, we find:

$\vec{d} \cdot \vec{c} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c}) = 0$,

which means $\vec{d}$ is perpendicular to $\vec{c}$.

Therefore, Statement II is correct.

∴ Both Statement I and Statement II are correct.

Hence, the correct answer is option 3.

Calculation:

Given,

$3\vec{a} - 4\vec{b} + \vec{c} = 0$

$\vec{c} = 4\vec{b} - 3\vec{a}$

The vector$\overrightarrow{AB}$ is:

$\overrightarrow{AB} = \vec{b} - \vec{a}$

The vector $\overrightarrow{BC}$ is:

$\overrightarrow{BC} = \vec{c} - \vec{b}$

Substituting $\vec{c} = 4\vec{b} - 3\vec{a}$:

$\overrightarrow{BC} = (4\vec{b} - 3\vec{a}) - \vec{b}$

$\overrightarrow{BC} = 3\vec{b} - 3\vec{a}$

Step 4: Now, $\overrightarrow{BC} = 3(\vec{b} - \vec{a})$, which gives:

$AB : BC = 1 : 3$

∴ The correct ratio is AB : BC = 1 : 3 , 

Hence, the correct answer is Option 2.

Calculation:

Given,

The position vectors of points A, B, and C are $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively, and $\vec{c} = \cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b}$.

The expression to evaluate is: $(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})$.

First, substitute $\vec{c}$ into the equation:

$(\vec{a} \times \vec{b}) + (\vec{b} \times (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b})) + (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b}) \times \vec{a}$.

Using the distributive property of the cross product:

$(\vec{a} \times \vec{b}) + \left[ (\vec{b} \times \cos^2 \theta , \vec{a}) + (\vec{b} \times \sin^2 \theta , \vec{b}) \right] + \left[ (\cos^2 \theta , \vec{a} \times \vec{a}) + (\sin^2 \theta , \vec{b} \times \vec{a}) \right]$.

Since $\vec{b} \times \vec{b} = 0$ and $\vec{a} \times \vec{a} = 0$, we are left with:

$(\vec{a} \times \vec{b}) + \cos^2 \theta , (\vec{b} \times \vec{a}) + \sin^2 \theta , (- \vec{a} \times \vec{b})$.

Substitute $\vec{b} \times \vec{a} = - (\vec{a} \times \vec{b})$ into the expression:

$(\vec{a} \times \vec{b}) + \cos^2 \theta , (- \vec{a} \times \vec{b}) + \sin^2 \theta , (- \vec{a} \times \vec{b})$.

Factor out $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} \left[ 1 - \cos^2 \theta - \sin^2 \theta \right]$.

Since $\cos^2 \theta + \sin^2 \theta = 1$, the expression becomes:

$\vec{a} \times \vec{b} [1 - 1] = 0$.

∴ The final result is $\vec{0}$.

Hence, the correct answer is option 1.

Calculation:

Given,

The vectors $\vec{a},\vec{b} ,(\vec{a}\times\vec{b})$ are unit vectors.

Since $\vec a$ and  $\vec{b}$ are unit vectors, we know:

$|\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1$.

The magnitude of the cross product $( \vec{a} \times \vec{b} )$ is given by:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \times 1 \times \sin \theta = \sin \theta$.

Since $( |\vec{a} \times \vec{b}| = 1$, we have:

$\sin \theta = 1$, so $\theta = 90^\circ$, meaning $\vec a$ and $\vec{b}$ are perpendicular.

The dot product $( \vec{a} \cdot \vec{b} )$ is:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \times 1 \times \cos 90^\circ = 0$.

∴ The value of$( \vec{a} \cdot \vec{b} )$ is 0.

Hence, the correct answer is Option 1.