For boron family (B,Al,Ga,In and Tl)
x : Number of elements which are solid at `40^(@)`
`y` : period number of element which has greater ionization energy than element just above and below it in periodic table.
` Z`: Period number of most abundant element of group `13`.
Report your answer `x+2y+3z`

Explanation

To solve the question step by step, we need to determine the values of \( x \), \( y \), and \( z \) based on the boron family (B, Al, Ga, In, Tl) and then compute \( x + 2y + 3z \).

### Step 1: Determine \( x \)
**Definition**: \( x \) is the number of elements in the boron family that are solid at \( 40^\circ C \).

- **Boron (B)**: Solid at \( 40^\circ C \).
- **Aluminum (Al)**: Solid at \( 40^\circ C \).
- **Gallium (Ga)**: Melts at \( 29.76^\circ C \), so it is not solid at \( 40^\circ C \).
- **Indium (In)**: Solid at \( 40^\circ C \).
- **Thallium (Tl)**: Solid at \( 40^\circ C \).

**Count**: The solid elements are B, Al, In, and Tl. Therefore, \( x = 4 \).

### Step 2: Determine \( y \)
**Definition**: \( y \) is the period number of the element that has a greater ionization energy than the elements just above and below it in the periodic table.

- **Ionization Energies**:
- Boron (B) > Aluminum (Al)
- Aluminum (Al) < Gallium (Ga)
- Gallium (Ga) < Indium (In)
- Indium (In) < Thallium (Tl)

From the above, we see that Boron (B) has a greater ionization energy than Aluminum (Al) and Gallium (Ga).

**Period Number**: Boron is in period 2, so \( y = 2 \).

### Step 3: Determine \( z \)
**Definition**: \( z \) is the period number of the most abundant element of group 13.

- The most abundant element in group 13 is Aluminum (Al).

**Period Number**: Aluminum is in period 3, so \( z = 3 \).

### Step 4: Calculate \( x + 2y + 3z \)
Now we can substitute the values we found into the expression:

\[ x + 2y + 3z = 4 + 2(2) + 3(3) \]

Calculating this step by step:

1. \( 2y = 2 \times 2 = 4 \)
2. \( 3z = 3 \times 3 = 9 \)
3. Now add them together: \( 4 + 4 + 9 = 17 \)

Thus, the final answer is:

\[ \boxed{17} \]