If lines `x+2y-1=0,a x+y+3=0,` and `b x-y+2=0` are concurrent, and `S` is the curve denoting the locus of `(a , b)` , then the least distance of `S` from the origin is `5/(sqrt(57))` (b) `5//sqrt(51)` `5//sqrt(58)` (d) `5//sqrt(59)`

Options

A. `5//sqrt(57)`
B. `5//sqrt(51)`
C. `5//sqrt(58)`
D. `5//sqrt(59)`

Explanation

To solve the problem step by step, we need to determine the condition for the three lines to be concurrent, find the locus of the parameters \(a\) and \(b\), and then calculate the least distance of this locus from the origin.

### Step 1: Write down the equations of the lines
The equations of the lines are given as:
1. \(x + 2y - 1 = 0\)
2. \(ax + y + 3 = 0\)
3. \(bx - y + 2 = 0\)

### Step 2: Set up the determinant for concurrency
For the three lines to be concurrent, the determinant of their coefficients must be zero. The coefficients of the lines can be arranged in a matrix as follows:

\[ \begin{vmatrix} 1 & 2 & -1 \\ a & 1 & 3 \\ b & -1 & 2 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ 1 \cdot \begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} a & 3 \\ b & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} a & 1 \\ b & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix} = (1 \cdot 2) - (3 \cdot -1) = 2 + 3 = 5\)

2. \(\begin{vmatrix} a & 3 \\ b & 2 \end{vmatrix} = (a \cdot 2) - (3 \cdot b) = 2a - 3b\)

3. \(\begin{vmatrix} a & 1 \\ b & -1 \end{vmatrix} = (a \cdot -1) - (1 \cdot b) = -a - b\)

Putting it all together, we have:

\[ 1 \cdot 5 - 2(2a - 3b) - 1(-a - b) = 0 \]

This simplifies to:

\[ 5 - 4a + 6b + a + b = 0 \]

Combining like terms gives:

\[ 7b - 3a + 5 = 0 \]

### Step 4: Rearranging the equation
We can rearrange this equation to express \(b\) in terms of \(a\):

\[ 7b = 3a - 5 \implies b = \frac{3a - 5}{7} \]

### Step 5: Identify the locus of \((a, b)\)
The equation \(b = \frac{3a - 5}{7}\) represents a straight line in the \(ab\)-plane.

### Step 6: Find the distance from the origin to the line
The line can be rewritten in standard form:

\[ 3a - 7b + 5 = 0 \]

Using the formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting \(A = 3\), \(B = -7\), \(C = 5\), and the point as the origin \((0, 0)\):

\[ d = \frac{|3(0) - 7(0) + 5|}{\sqrt{3^2 + (-7)^2}} = \frac{|5|}{\sqrt{9 + 49}} = \frac{5}{\sqrt{58}} \]

### Conclusion
Thus, the least distance of the locus \(S\) from the origin is:

\[ \frac{5}{\sqrt{58}} \]