Given:
– \(x + y + z = 6\sqrt{3}\)
– \(x^2 + y^2 + z^2 = 36\)

1. Use the identity \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\) to expand \((x + y + z)^2\):

\[ (x + y + z)^2 = (6\sqrt{3})^2 \]
\[ x^2 + y^2 + z^2 + 2(xy + yz + zx) = 108 \]
\[ 36 + 2(xy + yz + zx) = 108 \]
\[ 2(xy + yz + zx) = 72 \]
\[ xy + yz + zx = 36 \] (Equation A)

2. Comparing Equation A with \(x^2 + y^2 + z^2 = 36\):

\[ x^2 + y^2 + z^2 = xy + yz + zx \]

This implies that \(x = y = z\), as the sum of the squares of the variables is equal to the sum of their pairwise products.

3. Therefore, the ratio of \(x : y : z\) is \(1 : 1 : 1\).

Conclusion:
The ratio \(x : y : z\) is \(1 : 1 : 1\).

Given:
– We need to find the value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\).

1. We use the identity \(\cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x\):

\[ \cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x \]

2. Applying this identity to \(\cos 50^\circ, \cos 10^\circ, \cos 110^\circ\):

\[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = \frac{1}{4} \cos 150^\circ \]
\[ \cos 150^\circ = -\frac{\sqrt{3}}{2} \]
\[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = -\frac{\sqrt{3}}{8} \]

3. Also, \(\cos 60^\circ = \frac{1}{2}\).

4. Substituting these values into the given expression:

\[ \frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right) \]
\[ = \frac{16}{\sqrt{3}} \times \left(-\frac{\sqrt{3}}{8}\right) \times \frac{1}{2} \]
\[ = -1 \]

Conclusion:
The value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\) is \(-1\).

Given:
– \(x^4 + \frac{1}{x^4} = 322\) and \(x > 1\).

1. We know that \((a + b)^2 = a^2 + b^2 + 2ab\). Therefore, \((a + b)^2 – 2ab = a^2 + b^2\).

2. Applying this to \(x^2 + \frac{1}{x^2}\):
\[ \left(x^2 + \frac{1}{x^2}\right)^2 – 2 \times x^2 \times \frac{1}{x^2} = x^4 + \frac{1}{x^4} \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 322 + 2 \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 324 \]
\[ x^2 + \frac{1}{x^2} = \pm 18 \]
Since \(x > 1\), we take the positive root:
\[ x^2 + \frac{1}{x^2} = 18 \]

3. Similarly, for \(x^3 – \frac{1}{x^3}\), we use the identity \((a – b)^2 + 2ab = a^2 + b^2\):
\[ \left(x^2 – \frac{1}{x^2}\right)^2 + 2 = 18 \]
\[ \left(x^2 – \frac{1}{x^2}\right)^2 = 16 \]
\[ x^2 – \frac{1}{x^2} = \pm 4 \]
Since \(x > 1\), we take the positive root:
\[ x^2 – \frac{1}{x^2} = 4 \]

4. Now, cubing both sides:
\[ \left(x^3 – \frac{1}{x^3}\right) – 3 \times x \times \frac{1}{x} \left(x^2 – \frac{1}{x^2}\right) = 64 \]
\[ x^3 – \frac{1}{x^3} – 3 \times 4 = 64 \]
\[ x^3 – \frac{1}{x^3} = 64 + 12 \]
\[ x^3 – \frac{1}{x^3} = 76 \]

Conclusion:
The value of \(x^3 – \frac{1}{x^3}\) is 76.

Let’s calculate the percentage of land that the man can purchase with Rs. 2,50,000.

First, we’ll find out the total cost of the land:

Total cost of the land = Area of the land × Rate per square metre
= 10,000 m² × Rs. 2000/m²
= Rs. 2,00,00,000

Now, the man has Rs. 2,50,000 with him. The percentage of the land he can purchase with this amount is:

Percentage of land = (Amount he has / Total cost of the land) × 100
= (2,50,000 / 2,00,00,000) × 100
= 0.0125 × 100
= 1.25%

Therefore, the man can purchase 1.25% of the land with Rs. 2,50,000.

Given:
– Cashback offered is 15% on the marked price.
– Profit earned is 19% on the cost price.
– The difference between the cashback offered and the profit earned is Rs. 150.

1. Let the cost price of the article be Rs. \(x\).

2. The selling price (SP) of the article is 119% of the cost price (due to the 19% profit):
\[ \text{SP} = 1.19x \]

3. The marked price (MP) of the article is obtained by dividing the SP by \(1 – \text{cashback rate}\):
\[ \text{MP} = \frac{\text{SP}}{1 – 0.15} = \frac{1.19x}{0.85} = 1.4x \]

4. The difference between the cashback offered and the profit earned is given as Rs. 150:
\[ \text{Cashback offered} – \text{Profit earned} = 150 \]
\[ 1.4x \times 0.15 – 0.19x = 150 \]
\[ 0.21x – 0.19x = 150 \]
\[ 0.02x = 150 \]

5. Solving for \(x\):
\[ x = \frac{150}{0.02} = 7500 \]

Conclusion:
The cost price of the item is Rs. 7500.