`(a)/(b) : (c )/(d) : (e )/(f)=(2)/(3)`
`(2a+c+3e)/(2b+d+3f)=`
`(ma+nc+pe)/(mb+nd+pf)=`

Answer

To solve the problem step by step, we need to analyze the given ratios and expressions carefully.

### Step 1: Understand the Ratios
We are given the ratios:
\[ \frac{a}{b} : \frac{c}{d} : \frac{e}{f} = \frac{2}{3} \]
This means we can express \(a\), \(c\), and \(e\) in terms of \(b\), \(d\), and \(f\). Let’s assume:
- \(a = 2k\)
- \(b = 3k\)
- \(c = 2m\)
- \(d = 3m\)
- \(e = 2p\)
- \(f = 3p\)

### Step 2: Substitute in the Expression
Now, we need to find the value of:
\[ \frac{2a + c + 3e}{2b + d + 3f} \]
Substituting the values we have:
\[ \frac{2(2k) + (2m) + 3(2p)}{2(3k) + (3m) + 3(3p)} \]

### Step 3: Simplify the Numerator
Calculating the numerator:
\[ 2(2k) + (2m) + 3(2p) = 4k + 2m + 6p \]

### Step 4: Simplify the Denominator
Calculating the denominator:
\[ 2(3k) + (3m) + 3(3p) = 6k + 3m + 9p \]

### Step 5: Form the Final Expression
Now we can write the expression as:
\[ \frac{4k + 2m + 6p}{6k + 3m + 9p} \]

### Step 6: Factor Out Common Terms
We can factor out common terms from the numerator and denominator:
\[ = \frac{2(2k + m + 3p)}{3(2k + m + 3p)} \]

### Step 7: Cancel Common Factors
Since \(2k + m + 3p\) is common in both numerator and denominator, we can cancel it out:
\[ = \frac{2}{3} \]

### Step 8: Find the Second Expression
Now we need to find the value of:
\[ \frac{ma + nc + pe}{mb + nd + pf} \]
Substituting the values:
\[ \frac{m(2k) + n(2m) + p(2p)}{m(3k) + n(3m) + p(3p)} \]

### Step 9: Simplify the Second Expression
This simplifies to:
\[ \frac{2mk + 2nm + 2p}{3mk + 3nm + 3p} \]

### Step 10: Factor Out Common Terms Again
Factoring out the common term \(2\) in the numerator and \(3\) in the denominator gives:
\[ = \frac{2(mk + nm + p)}{3(mk + nm + p)} \]

### Step 11: Cancel Common Factors Again
Again, we can cancel \(mk + nm + p\):
\[ = \frac{2}{3} \]

### Final Result
Thus, the final answer for both expressions is:
\[ \frac{2}{3} \]