Find the equation of the plane through the line of intersection of :
`vec(r). (2 hat(i) - 3hat(j) + 4 hat(k) ) = 1 and vec(r). (hat(i) - hat(j) ) + 4 = 0`
and perpendicular to the plane ` vec(r). (2 hat(i) - hat(j) + hat(k)) + 8 = 0`.
hence, find whether the plane thus obtained contains the line:
x - 1 = 2y - 4 = 3z - 12.

Answer

To find the equation of the plane through the line of intersection of the given planes and perpendicular to another plane, we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes are:
1. \( \vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 1 \)
2. \( \vec{r} \cdot (\hat{i} - \hat{j}) + 4 = 0 \)

The equation of the plane that is perpendicular to another plane is given by:
3. \( \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) + 8 = 0 \)

### Step 2: Write the general equation of the required plane
The general equation of the plane through the line of intersection of the first two planes can be expressed as:
\[ \vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) - 1 + \lambda (\vec{r} \cdot (\hat{i} - \hat{j}) + 4) = 0 \]
where \( \lambda \) is a scalar.

### Step 3: Simplify the equation
Expanding this, we have:
\[ \vec{r} \cdot (2 + \lambda) \hat{i} + \vec{r} \cdot (-3 - \lambda) \hat{j} + \vec{r} \cdot (4) \hat{k} - 1 + 4\lambda = 0 \]
This can be rewritten as:
\[ \vec{r} \cdot ((2 + \lambda) \hat{i} + (-3 - \lambda) \hat{j} + 4 \hat{k}) = 1 - 4\lambda \]

### Step 4: Find the direction ratios
The direction ratios of the plane can be denoted as:
\[ (2 + \lambda, -3 - \lambda, 4) \]

### Step 5: Use the perpendicular condition
The plane is perpendicular to the plane given by:
\[ \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) + 8 = 0 \]
The direction ratios of this plane are \( (2, -1, 1) \).

For the two planes to be perpendicular, their direction ratios must satisfy:
\[ (2 + \lambda) \cdot 2 + (-3 - \lambda)(-1) + 4 \cdot 1 = 0 \]
Expanding this gives:
\[ 4 + 2\lambda + 3 + \lambda + 4 = 0 \]
which simplifies to:
\[ 3\lambda + 11 = 0 \]
Thus, solving for \( \lambda \):
\[ \lambda = -\frac{11}{3} \]

### Step 6: Substitute \( \lambda \) back into the equation
Substituting \( \lambda \) back into the equation of the plane:
\[ \vec{r} \cdot \left(2 - \frac{11}{3}\right) \hat{i} + \vec{r} \cdot \left(-3 + \frac{11}{3}\right) \hat{j} + \vec{r} \cdot (4) \hat{k} = 1 + \frac{44}{3} \]
Calculating each component:
- For \( \hat{i} \): \( 2 - \frac{11}{3} = \frac{6 - 11}{3} = -\frac{5}{3} \)
- For \( \hat{j} \): \( -3 + \frac{11}{3} = -\frac{9}{3} + \frac{11}{3} = \frac{2}{3} \)
- The constant on the right side becomes \( \frac{3 + 44}{3} = \frac{47}{3} \)

Thus, the equation of the plane becomes:
\[ \vec{r} \cdot \left(-\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + 4 \hat{k}\right) = \frac{47}{3} \]

### Step 7: Convert to standard form
Multiplying through by 3 to eliminate the fraction gives:
\[ -5x + 2y + 12z - 47 = 0 \]

### Step 8: Check if the plane contains the line
The line is given by:
\[ x - 1 = 2y - 4 = 3z - 12 \]
This can be rewritten in parametric form:
- \( x = 1 + t \)
- \( y = 2 + \frac{t}{2} \)
- \( z = 4 + t \)

To check if the plane contains this line, substitute the parametric equations into the plane equation:
\[ -5(1 + t) + 2\left(2 + \frac{t}{2}\right) + 12(4 + t) - 47 = 0 \]
Simplifying this will show if the left-hand side equals zero for all \( t \).

### Conclusion
If the distance calculated is zero, then the line lies on the plane. Thus, the required plane contains the given line.