Taking horizontal ground as `xz`- plane and y-axis vertically upwards. A particle is projected from the horizontal ground from a point whose position vector is `hati + hatk` meter with initial velocity 2i +3j+k from the origin when it hits ground is given by `(1)/(5)[11hati+n hatk]`, find `n`. (Take `g = 10 m//s^(2)` in negative y - direction)

Explanation

To solve the problem, we need to analyze the motion of the particle projected from a point in the xz-plane with an initial velocity. We will use the equations of motion to find the value of \( n \).

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The particle is projected from the point with position vector \( \mathbf{r_1} = \hat{i} + \hat{k} \) meters.
- The initial velocity \( \mathbf{u} = 2\hat{i} + 3\hat{j} + \hat{k} \) m/s.
- The acceleration due to gravity \( \mathbf{a} = 0\hat{i} - 10\hat{j} + 0\hat{k} \) m/s² (acting in the negative y-direction).

2. **Use the Equation of Motion:**
The position vector \( \mathbf{r} \) at time \( t \) can be expressed as:
\[ \mathbf{r} = \mathbf{r_1} + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \]
Substituting the known values:
\[ \mathbf{r} = (\hat{i} + \hat{k}) + (2\hat{i} + 3\hat{j} + \hat{k})t + \frac{1}{2}(0\hat{i} - 10\hat{j} + 0\hat{k})t^2 \]

3. **Simplify the Position Vector:**
\[ \mathbf{r} = \hat{i} + \hat{k} + (2t\hat{i} + 3t\hat{j} + t\hat{k}) - 5t^2\hat{j} \]
\[ \mathbf{r} = (1 + 2t)\hat{i} + (3t - 5t^2)\hat{j} + (1 + t)\hat{k} \]

4. **Condition for Hitting the Ground:**
The particle hits the ground when the y-component of the position vector is zero:
\[ 3t - 5t^2 = 0 \]
Factoring out \( t \):
\[ t(3 - 5t) = 0 \]
This gives us \( t = 0 \) (initial time) or \( t = \frac{3}{5} \) seconds.

5. **Find the Position Vector When it Hits the Ground:**
Substitute \( t = \frac{3}{5} \) into the position vector \( \mathbf{r} \):
\[ \mathbf{r} = (1 + 2 \cdot \frac{3}{5})\hat{i} + (3 \cdot \frac{3}{5} - 5 \cdot (\frac{3}{5})^2)\hat{j} + (1 + \frac{3}{5})\hat{k} \]
Calculate each component:
- For \( x \):
\[ x = 1 + \frac{6}{5} = \frac{11}{5} \]
- For \( y \):
\[ y = 0 \quad \text{(since it hits the ground)} \]
- For \( z \):
\[ z = 1 + \frac{3}{5} = \frac{8}{5} \]

6. **Final Position Vector:**
Thus, the position vector when it hits the ground is:
\[ \mathbf{r} = \frac{1}{5}(11\hat{i} + n\hat{k}) \]
Comparing with our calculated position vector:
\[ \frac{1}{5}(11\hat{i} + n\hat{k}) = \frac{11}{5}\hat{i} + \frac{8}{5}\hat{k} \]
This gives us \( n = 8 \).

### Conclusion:
The value of \( n \) is \( 8 \).