Equation of the plane passing through the point `(1, -1, 3)`, parallel to the vector `hati+2hatj+4hatk` and perependicular to the plane `x-2y+z=6` is given by `ax+by+cz+8=0`, then the value of `2a-5b+7c` is equal to

Options

A. 32
B. 31
C. `-(184)/(5)`
D. `(72)/(5)`

Answer

To find the equation of the plane that passes through the point \( (1, -1, 3) \), is parallel to the vector \( \hat{i} + 2\hat{j} + 4\hat{k} \), and is perpendicular to the plane given by the equation \( x - 2y + z = 6 \), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The normal vector of the plane \( x - 2y + z = 6 \) can be extracted from its coefficients. Thus, the normal vector \( \vec{n_1} \) is:
\[ \vec{n_1} = \langle 1, -2, 1 \rangle \]

### Step 2: Identify the direction vector of the required plane
The required plane is parallel to the vector \( \hat{i} + 2\hat{j} + 4\hat{k} \). We denote this vector as:
\[ \vec{v} = \langle 1, 2, 4 \rangle \]

### Step 3: Find the normal vector of the required plane
Since the required plane is perpendicular to the given plane, the normal vector \( \vec{n_2} \) of the required plane can be found using the cross product of the normal vector of the given plane \( \vec{n_1} \) and the direction vector \( \vec{v} \):
\[ \vec{n_2} = \vec{n_1} \times \vec{v} \]

Calculating the cross product:
\[ \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & 4 \end{vmatrix} = \hat{i}((-2)(4) - (1)(2)) - \hat{j}((1)(4) - (1)(1)) + \hat{k}((1)(2) - (1)(-2)) \]
\[ = \hat{i}(-8 - 2) - \hat{j}(4 - 1) + \hat{k}(2 + 2) \]
\[ = -10\hat{i} - 3\hat{j} + 4\hat{k} \]
Thus, the normal vector of the required plane is:
\[ \vec{n_2} = \langle -10, -3, 4 \rangle \]

### Step 4: Write the equation of the plane
The equation of a plane can be written in the form:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( \langle a, b, c \rangle \) is the normal vector. Substituting the point \( (1, -1, 3) \) and the normal vector \( \langle -10, -3, 4 \rangle \):
\[ -10(x - 1) - 3(y + 1) + 4(z - 3) = 0 \]
Expanding this:
\[ -10x + 10 - 3y - 3 + 4z - 12 = 0 \]
\[ -10x - 3y + 4z - 5 = 0 \]
Rearranging gives:
\[ 10x + 3y - 4z + 5 = 0 \]

### Step 5: Compare with the given equation
The equation is given in the form \( ax + by + cz + 8 = 0 \). We can rewrite our equation as:
\[ 10x + 3y - 4z + 5 = 0 \]
This implies:
\[ a = 10, \quad b = 3, \quad c = -4 \]

### Step 6: Calculate \( 2a - 5b + 7c \)
Now we substitute the values of \( a \), \( b \), and \( c \):
\[ 2a - 5b + 7c = 2(10) - 5(3) + 7(-4) \]
\[ = 20 - 15 - 28 \]
\[ = 20 - 15 - 28 = -23 \]

### Final Answer
Thus, the value of \( 2a - 5b + 7c \) is:
\[ \boxed{-23} \]