To find the values of \(\alpha\) and \(\beta\) given that the cross product of two vectors is the zero vector, we need to apply the fundamental property of vector cross products.

Vector Cross Product Principle

The cross product of two non-zero vectors, let's denote them as \(\vec{A}\) and \(\vec{B}\), yields the zero vector \(\overrightarrow{0}\) if and only if the two vectors are parallel or collinear. This is a crucial concept in vector algebra. When two vectors are parallel, it means that one vector is simply a scalar multiple of the other. Consequently, their corresponding components are proportional.

If we have two vectors \(\vec{A} = A_x \hat{\imath} + A_y \hat{\jmath} + A_z \hat{k}\) and \(\vec{B} = B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k}\), and they are parallel, then the following relationship holds true:

\[ \frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z} = k \]

where \(k\) is a scalar constant representing the ratio of their magnitudes.

In our given problem, the two vectors are:

• First vector: \(\vec{V_1} = 2 \hat{\imath}+6 \hat{\jmath}+27 \hat{k}\)
• Second vector: \(\vec{V_2} = \hat{\imath}+\alpha \hat{\jmath}+\beta \hat{k}\)

The problem states that their cross product is the zero vector:

\[ (2 \hat{\imath}+6 \hat{\jmath}+27 \hat{k}) \times (\hat{\imath}+\alpha \hat{\jmath}+\beta \hat{k})=\overrightarrow{0} \]

Figure-out Alpha and Beta Values

Since the cross product of \(\vec{V_1}\) and \(\vec{V_2}\) is \(\overrightarrow{0}\), it directly implies that \(\vec{V_1}\) and \(\vec{V_2}\) are parallel vectors. Therefore, their corresponding components must be in proportion. We can set up the proportionality equation by comparing the coefficients of \(\hat{\imath}\), \(\hat{\jmath}\), and \(\hat{k}\) from both vectors:

\[ \frac{\text{Coefficient of } \hat{\imath} \text{ in } \vec{V_1}}{\text{Coefficient of } \hat{\imath} \text{ in } \vec{V_2}} = \frac{\text{Coefficient of } \hat{\jmath} \text{ in } \vec{V_1}}{\text{Coefficient of } \hat{\jmath} \text{ in } \vec{V_2}} = \frac{\text{Coefficient of } \hat{k} \text{ in } \vec{V_1}}{\text{Coefficient of } \hat{k} \text{ in } \vec{V_2}} \]

Plugging in the given values, we get:

\[ \frac{2}{1} = \frac{6}{\alpha} = \frac{27}{\beta} \]

Solving for Alpha (\(\alpha\))

Let's take the first two parts of the equality to solve for \(\alpha\):

\[ \frac{2}{1} = \frac{6}{\alpha} \]

To find \(\alpha\), we can cross-multiply:

\[ 2 \times \alpha = 1 \times 6 \]

\[ 2\alpha = 6 \]

Now, divide both sides by 2:

\[ \alpha = \frac{6}{2} \]

\[ \alpha = 3 \]

Solving for Beta (\(\beta\))

Now, let's use the first and the third parts of the equality to solve for \(\beta\):

\[ \frac{2}{1} = \frac{27}{\beta} \]

Again, cross-multiply to solve for \(\beta\):

\[ 2 \times \beta = 1 \times 27 \]

\[ 2\beta = 27 \]

Divide both sides by 2:

\[ \beta = \frac{27}{2} \]

Values Verification

Based on our calculations, the values for \(\alpha\) and \(\beta\) are:

• \(\alpha = 3\)
• \(\beta = \frac{27}{2}\)

We can summarize our findings in the table below:

These calculated values match one of the provided options, confirming our solution for \(\alpha\) and \(\beta\).