A vector `veca=(x,y,z)` makes an obtuse angle with F-axis, and make equal angles with `vecb=(y,-2z, 3x)` and `vecc=(2z, 3x,-y)` and `veca` is perpendicular to `vecd = (1,-1,2)` if `|veca|=2sqrt(3)` then vector `veca` is:

Options

A. (1,2,3)
B. (2,-2,-2)
C. (-1,2,4)
D. None of these

Answer

To solve the problem step by step, we will analyze the given conditions and derive the required vector \(\vec{a} = (x, y, z)\).

### Step 1: Analyze the conditions
We know:
1. \(\vec{a}\) makes an obtuse angle with the F-axis, which implies that the \(z\) component must be negative (since the F-axis is aligned with the positive \(z\)-axis).
2. \(\vec{a}\) makes equal angles with \(\vec{b} = (y, -2z, 3x)\) and \(\vec{c} = (2z, 3x, -y)\).
3. \(\vec{a}\) is perpendicular to \(\vec{d} = (1, -1, 2)\).
4. The magnitude of \(\vec{a}\) is given as \(|\vec{a}| = 2\sqrt{3}\).

### Step 2: Set up the equations
From the condition that \(\vec{a}\) is perpendicular to \(\vec{d}\):
\[ \vec{a} \cdot \vec{d} = 0 \]
This gives us:
\[ x - y + 2z = 0 \quad \text{(Equation 1)} \]

### Step 3: Equal angles condition
Since \(\vec{a}\) makes equal angles with \(\vec{b}\) and \(\vec{c}\), we can use the dot product:
\[ \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \]
Calculating \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = x \cdot y + y \cdot (-2z) + z \cdot (3x) = xy - 2yz + 3xz \]
Calculating \(\vec{a} \cdot \vec{c}\):
\[ \vec{a} \cdot \vec{c} = x \cdot (2z) + y \cdot (3x) + z \cdot (-y) = 2xz + 3xy - yz \]
Setting these equal:
\[ xy - 2yz + 3xz = 2xz + 3xy - yz \]
Rearranging gives:
\[ -2yz + 3xz - 2xz - 3xy + yz = 0 \]
This simplifies to:
\[ xz - 2xy - yz = 0 \quad \text{(Equation 2)} \]

### Step 4: Magnitude condition
The magnitude condition gives us:
\[ x^2 + y^2 + z^2 = (2\sqrt{3})^2 = 12 \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
From Equation 1:
\[ x = y - 2z \]
Substituting \(x\) into Equation 2:
\[ (y - 2z)z - 2(y - 2z)y - yz = 0 \]
Expanding and simplifying:
\[ yz - 2z^2 - 2y^2 + 4zy - yz = 0 \]
This simplifies to:
\[ -2z^2 + 4zy - 2y^2 = 0 \]
Factoring out:
\[ -2(z^2 - 2zy + y^2) = 0 \]
This gives:
\[ z^2 - 2zy + y^2 = 0 \implies (z - y)^2 = 0 \implies z = y \]
Substituting \(z = y\) into Equation 1:
\[ x = y - 2y = -y \]
Substituting \(x = -y\) and \(z = y\) into Equation 3:
\[ (-y)^2 + y^2 + y^2 = 12 \implies 3y^2 = 12 \implies y^2 = 4 \implies y = 2 \text{ or } y = -2 \]
Since \(z\) is negative, we take \(y = -2\) and thus \(z = -2\) and \(x = 2\).

### Final Vector
The vector \(\vec{a}\) is:
\[ \vec{a} = (2, -2, -2) \]