Find the equation of the plane through the line of intersection of the planes `vecr.(2hati-3hatj+4hatk)=1` and `vecr.(hati-hatj)+4=0` and perpendicular to the plane `vecr.(2hati-hatj+hatk)+8=0`.

Answer

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

### Step 1: Write the equations of the given planes in Cartesian form.

The equations of the planes are given in vector form:
1. Plane 1: \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 1\)
2. Plane 2: \(\vec{r} \cdot (\hat{i} - \hat{j}) + 4 = 0\)

Converting these to Cartesian form:
1. For Plane 1, we have:
\[ 2x - 3y + 4z - 1 = 0 \]
2. For Plane 2, we have:
\[ x - y + 4 = 0 \quad \Rightarrow \quad x - y + 4 = 0 \]

### Step 2: Form the general equation of the plane through the line of intersection.

The general equation of the plane through the intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of Plane 1 and Plane 2:
\[ (2x - 3y + 4z - 1) + \lambda (x - y + 4) = 0 \]

### Step 3: Combine and simplify the equation.

Expanding and combining like terms gives:
\[ (2 + \lambda)x + (-3 - \lambda)y + 4z + (4\lambda - 1) = 0 \]
Thus, the equation of the required plane (let's call it Plane 3) is:
\[ (2 + \lambda)x + (-3 - \lambda)y + 4z + (4\lambda - 1) = 0 \]

### Step 4: Find the normal vector of Plane 3.

The normal vector \( \vec{n_3} \) of Plane 3 is:
\[ \vec{n_3} = (2 + \lambda, -3 - \lambda, 4) \]

### Step 5: Write the equation of the plane perpendicular to Plane 4.

The equation of Plane 4 is given as:
\[ \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 8 = 0 \]
Thus, the normal vector \( \vec{n_4} \) of Plane 4 is:
\[ \vec{n_4} = (2, -1, 1) \]

### Step 6: Set up the condition for perpendicularity.

For the planes to be perpendicular, the dot product of their normal vectors must be zero:
\[ \vec{n_3} \cdot \vec{n_4} = 0 \]
Substituting the normal vectors:
\[ (2 + \lambda) \cdot 2 + (-3 - \lambda) \cdot (-1) + 4 \cdot 1 = 0 \]
Expanding this gives:
\[ 2(2 + \lambda) + (3 + \lambda) + 4 = 0 \]
\[ 4 + 2\lambda + 3 + \lambda + 4 = 0 \]
\[ (3\lambda + 11) = 0 \]
Solving for \( \lambda \):
\[ 3\lambda = -11 \quad \Rightarrow \quad \lambda = -\frac{11}{3} \]

### Step 7: Substitute \( \lambda \) back into the equation of Plane 3.

Now substitute \( \lambda = -\frac{11}{3} \) into the equation of Plane 3:
\[ (2 - \frac{11}{3})x + (-3 + \frac{11}{3})y + 4z + (4 \cdot -\frac{11}{3} - 1) = 0 \]
Calculating each term:
1. \( 2 - \frac{11}{3} = \frac{6}{3} - \frac{11}{3} = -\frac{5}{3} \)
2. \( -3 + \frac{11}{3} = -\frac{9}{3} + \frac{11}{3} = \frac{2}{3} \)
3. \( 4 \cdot -\frac{11}{3} - 1 = -\frac{44}{3} - \frac{3}{3} = -\frac{47}{3} \)

Thus, the equation becomes:
\[ -\frac{5}{3}x + \frac{2}{3}y + 4z - \frac{47}{3} = 0 \]

### Step 8: Multiply through by 3 to eliminate fractions.

Multiplying the entire equation by 3 gives:
\[ -5x + 2y + 12z - 47 = 0 \]
Rearranging gives the final equation of the required plane:
\[ -5x + 2y + 12z = 47 \]

### Final Answer:
The equation of the required plane is:
\[ -5x + 2y + 12z = 47 \]