If a, b, c, are non-coplanar vectors such that `(2h+k)a+(3-4h+l)b+(1+h+k)c=ha+kb+lc` then

Options

A. `h = 1, k = -4//3, l = 4//3`
B. `h=4//3, k=-4//3,l=1`
C. `h=1//3,k=-1//3, l=2//3`
D. none of these

Answer

To solve the equation given in the question, we need to compare the coefficients of the vectors \( a \), \( b \), and \( c \) from both sides of the equation. The equation given is:

\[ (2h+k)a + (3-4h+l)b + (1+h+k)c = ha + kb + lc \]

### Step 1: Compare Coefficients
We can equate the coefficients of \( a \), \( b \), and \( c \) from both sides of the equation.

1. Coefficient of \( a \):
\[ 2h + k = h \tag{1} \]

2. Coefficient of \( b \):
\[ 3 - 4h + l = k \tag{2} \]

3. Coefficient of \( c \):
\[ 1 + h + k = l \tag{3} \]

### Step 2: Solve the First Equation
From equation (1):
\[ 2h + k = h \]
Rearranging gives:
\[ 2h - h + k = 0 \implies h + k = 0 \implies k = -h \tag{4} \]

### Step 3: Substitute \( k \) in the Second Equation
Now substitute \( k = -h \) into equation (2):
\[ 3 - 4h + l = -h \]
Rearranging gives:
\[ 3 - 4h + l + h = 0 \implies 3 - 3h + l = 0 \implies l = 3h - 3 \tag{5} \]

### Step 4: Substitute \( k \) and \( l \) in the Third Equation
Now substitute \( k = -h \) and \( l = 3h - 3 \) into equation (3):
\[ 1 + h + (-h) = 3h - 3 \]
This simplifies to:
\[ 1 = 3h - 3 \]
Rearranging gives:
\[ 3h = 4 \implies h = \frac{4}{3} \tag{6} \]

### Step 5: Find \( k \) and \( l \)
Now that we have \( h \), we can find \( k \) and \( l \).

1. From equation (4):
\[ k = -h = -\frac{4}{3} \tag{7} \]

2. From equation (5):
\[ l = 3h - 3 = 3 \cdot \frac{4}{3} - 3 = 4 - 3 = 1 \tag{8} \]

### Final Values
Thus, we have:
\[ h = \frac{4}{3}, \quad k = -\frac{4}{3}, \quad l = 1 \]

### Summary
The values of \( h \), \( k \), and \( l \) are:
- \( h = \frac{4}{3} \)
- \( k = -\frac{4}{3} \)
- \( l = 1 \)