The portion of the line `x+3y-1 =0` intersepted between the lines `ax+y+1 = 0` and `x + 3y =0` subtend a right angle at origin , then the value of |a| is

Answer

To solve the problem, we need to find the value of |a| such that the portion of the line \( x + 3y - 1 = 0 \) intercepted between the lines \( ax + y + 1 = 0 \) and \( x + 3y = 0 \) subtends a right angle at the origin.

### Step 1: Find the intersection points of the lines

We start by finding the intersection points of the lines \( ax + y + 1 = 0 \) and \( x + 3y = 0 \).

1. Rearranging \( x + 3y = 0 \) gives us \( y = -\frac{1}{3}x \).
2. Substitute \( y \) into the first equation:
\[ ax + \left(-\frac{1}{3}x\right) + 1 = 0 \]
\[ (a - \frac{1}{3})x + 1 = 0 \]
\[ (a - \frac{1}{3})x = -1 \]
\[ x = \frac{-1}{a - \frac{1}{3}} \quad \text{(valid if } a \neq \frac{1}{3}\text{)} \]

3. Now, substituting \( x \) back to find \( y \):
\[ y = -\frac{1}{3} \left(\frac{-1}{a - \frac{1}{3}}\right) = \frac{1}{3(a - \frac{1}{3})} \]

Thus, the intersection point \( P \) is:
\[ P\left(\frac{-1}{a - \frac{1}{3}}, \frac{1}{3(a - \frac{1}{3})}\right) \]

### Step 2: Find the slope of the line segment OP

The slope \( m_1 \) of the line segment \( OP \) (from the origin to point \( P \)) is given by:
\[ m_1 = \frac{y}{x} = \frac{\frac{1}{3(a - \frac{1}{3})}}{\frac{-1}{a - \frac{1}{3}}} = -\frac{1}{3} \]

### Step 3: Find the slope of the line \( x + 3y - 1 = 0 \)

Rearranging the line \( x + 3y - 1 = 0 \):
\[ 3y = -x + 1 \implies y = -\frac{1}{3}x + \frac{1}{3} \]
Thus, the slope \( m_2 \) of this line is:
\[ m_2 = -\frac{1}{3} \]

### Step 4: Condition for right angle

For the lines to subtend a right angle at the origin, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ -\frac{1}{3} \cdot \left(-\frac{1}{3}\right) = \frac{1}{9} \neq -1 \]
This indicates that we must consider the slopes of the lines \( ax + y + 1 = 0 \) and \( x + 3y = 0 \).

### Step 5: Find the slope of line \( ax + y + 1 = 0 \)

Rearranging gives:
\[ y = -ax - 1 \]
Thus, the slope \( m_3 \) is:
\[ m_3 = -a \]

### Step 6: Apply the right angle condition

Now, we need the product of the slopes \( m_2 \) and \( m_3 \) to equal -1:
\[ -\frac{1}{3} \cdot (-a) = -1 \]
This simplifies to:
\[ \frac{a}{3} = -1 \implies a = -3 \]

### Step 7: Find the value of |a|

Thus, the value of \( |a| \) is:
\[ |a| = |-3| = 3 \]

### Final Answer:
\[ \boxed{3} \]