Let `veca` be a unit vector coplanar with `hati-hatj+2hatk` and `2hati-hatj+hatk` such that `veca` is perpendicular to `hati-2hatj+hatk`. If the projecton of `veca` along `hati-hatj+2hatk` is `lambda`, then the value of `(1)/(lambda^(2))` is equal to

Answer

To solve the problem step-by-step, we will follow the instructions provided in the video transcript and clarify each step.

### Step 1: Define the vectors
Let:
- \(\vec{a}\) be the unit vector we need to find.
- \(\vec{\alpha} = \hat{i} - \hat{j} + 2\hat{k}\)
- \(\vec{\beta} = 2\hat{i} - \hat{j} + \hat{k}\)
- \(\vec{\gamma} = \hat{i} - 2\hat{j} + \hat{k}\)

### Step 2: Calculate the dot products
We need to find the dot products of \(\vec{\gamma}\) with \(\vec{\alpha}\) and \(\vec{\beta}\).

1. **Calculate \(\vec{\gamma} \cdot \vec{\beta}\)**:
\[ \vec{\gamma} \cdot \vec{\beta} = (1)(2) + (-2)(-1) + (1)(1) = 2 + 2 + 1 = 5 \]

2. **Calculate \(\vec{\gamma} \cdot \vec{\alpha}\)**:
\[ \vec{\gamma} \cdot \vec{\alpha} = (1)(1) + (-2)(-1) + (1)(2) = 1 + 2 + 2 = 5 \]

### Step 3: Formulate the expression for \(\vec{a}\)
Using the results from the dot products, we can express \(\vec{a}\) as:
\[ \vec{a} = \frac{5\vec{\alpha} - 5\vec{\beta}}{||5\vec{\alpha} - 5\vec{\beta}||} \]

### Step 4: Calculate \(5\vec{\alpha} - 5\vec{\beta}\)
First, calculate \(\vec{\alpha} - \vec{\beta}\):
\[ \vec{\alpha} - \vec{\beta} = (\hat{i} - \hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 0\hat{j} + \hat{k} = -\hat{i} + \hat{k} \]
Thus,
\[ 5(\vec{\alpha} - \vec{\beta}) = 5(-\hat{i} + \hat{k}) = -5\hat{i} + 5\hat{k} \]

### Step 5: Calculate the magnitude
Now we need to find the magnitude of \(-5\hat{i} + 5\hat{k}\):
\[ ||-5\hat{i} + 5\hat{k}|| = \sqrt{(-5)^2 + 0^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 6: Write the unit vector \(\vec{a}\)
Now we can write \(\vec{a}\):
\[ \vec{a} = \frac{-5\hat{i} + 5\hat{k}}{5\sqrt{2}} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}} \]

### Step 7: Calculate the projection of \(\vec{a}\) on \(\vec{\alpha}\)
The projection of \(\vec{a}\) on \(\vec{\alpha}\) is given by:
\[ \text{proj}_{\vec{\alpha}} \vec{a} = \frac{\vec{a} \cdot \vec{\alpha}}{||\vec{\alpha}||} \]

1. **Calculate \(\vec{a} \cdot \vec{\alpha}\)**:
\[ \vec{a} \cdot \vec{\alpha} = \left(\frac{-1}{\sqrt{2}}\right)(1) + \left(0\right)(-1) + \left(\frac{1}{\sqrt{2}}\right)(2) = \frac{-1}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

2. **Calculate \(||\vec{\alpha}||\)**:
\[ ||\vec{\alpha}|| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

3. **Calculate the projection**:
\[ \text{proj}_{\vec{\alpha}} \vec{a} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{6}} = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

### Step 8: Find \(\lambda\) and calculate \(\frac{1}{\lambda^2}\)
Let \(\lambda = \frac{1}{\sqrt{12}}\). Therefore:
\[ \lambda^2 = \frac{1}{12} \]
Thus,
\[ \frac{1}{\lambda^2} = 12 \]

### Final Answer
The value of \(\frac{1}{\lambda^2}\) is \(12\).