Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
the system has infinite solutions if (a) `lambda ne 3` (b) `lambda =3,mu =10` (c) `lambda =3,mu ne 10` (d) `lambda =3,mu ne 10`

Options

A. `lambda ne 3`
B. `lambda =3,mu =10`
C. `lambda =3,mu ne 10`
D. `lambda =3,mu ne 10`

Explanation

To determine the conditions under which the given system of equations has infinite solutions, we need to analyze the equations provided:

1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 10 \) (Equation 2)
3. \( x + 2y + \lambda z = \mu \) (Equation 3)

### Step 1: Identify the coefficients
For the system to have infinite solutions, the equations must represent the same plane or be parallel. This occurs when the ratios of the coefficients of the variables in the equations are equal.

From the equations, we can extract the coefficients:
- For Equation 1: Coefficients are \( (1, 1, 1) \)
- For Equation 2: Coefficients are \( (1, 2, 3) \)
- For Equation 3: Coefficients are \( (1, 2, \lambda) \)

### Step 2: Set up the ratio of coefficients
To find the condition for infinite solutions, we need to compare the ratios of the coefficients of the equations:

For Equations 2 and 3:
\[ \frac{1}{1} = \frac{2}{2} = \frac{3}{\lambda} \]

From the first two ratios, we have:
\[ 1 = 1 \quad \text{(which is always true)} \]
From the last ratio, we need:
\[ \frac{3}{\lambda} = 1 \implies \lambda = 3 \]

### Step 3: Analyze the constant terms
Next, we need to ensure that the constant terms also satisfy the condition for infinite solutions. For Equations 2 and 3, we compare the constant terms:
\[ \frac{10}{\mu} = 1 \]
This implies:
\[ \mu = 10 \]

### Conclusion
Thus, for the system of equations to have infinite solutions, the conditions we derived are:
- \( \lambda = 3 \)
- \( \mu = 10 \)

### Final Answer
The correct option is (b) \( \lambda = 3, \mu = 10 \).