Any first degree equation in x, y and z represents a plane i.e., `ax+by+cz+d=0` is the general equation of a plane. If p be the length of perpendicular from the origin to a plane and d.c. of this normal is `lt l, m n gt`, then the equation of the plane in the normal form is `lx+my+nz=p_(1)`.
Vector equation of a plane passing through a point having position vector `vec(a)` and normal to vector `vec(n)` is
`(vec(r)-vec(a))vec(n)=0" or "vec(r).vec(n)=vec(a).vec(n)`
Suppose a vector `vec(n)` of magnitude `2sqrt(3)` such that it makes equal acual angles with the co-ordinate axes. If `vec(n)` is a normal to the plane containing the point `(1,-1,2)`.
The vector `vec(n)` is equal to

Options

A. `hati+hatj+hatk`
B. `2(hati+hatj+hatk)`
C. `hati-hatj+hatk`
D. None of these

Answer

To solve the problem, we need to find the vector \(\vec{n}\) that is normal to the plane and has a magnitude of \(2\sqrt{3}\), while also making equal angles with the coordinate axes.

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
Since the vector \(\vec{n}\) makes equal angles with the coordinate axes, we can denote the direction cosines as:
\[ l = m = n = \cos(\alpha) \]
where \(\alpha\) is the angle that the vector makes with each axis.

2. **Using the Relation of Direction Cosines**:
We know that the sum of the squares of the direction cosines equals 1:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting \(l = m = n = \cos(\alpha)\):
\[ 3\cos^2(\alpha) = 1 \]
Therefore,
\[ \cos^2(\alpha) = \frac{1}{3} \]
Taking the square root gives:
\[ \cos(\alpha) = \pm \frac{1}{\sqrt{3}} \]

3. **Finding the Vector \(\vec{n}\)**:
The vector \(\vec{n}\) can be expressed in terms of its direction cosines:
\[ \vec{n} = l\hat{i} + m\hat{j} + n\hat{k} = \cos(\alpha)\hat{i} + \cos(\alpha)\hat{j} + \cos(\alpha)\hat{k} \]
Substituting the value of \(\cos(\alpha)\):
\[ \vec{n} = \pm \frac{1}{\sqrt{3}} \hat{i} + \pm \frac{1}{\sqrt{3}} \hat{j} + \pm \frac{1}{\sqrt{3}} \hat{k} \]

4. **Magnitude of the Vector**:
The magnitude of the vector \(\vec{n}\) is given as \(2\sqrt{3}\). Thus, we can express the vector in terms of its magnitude and direction cosines:
\[ |\vec{n}| = \sqrt{l^2 + m^2 + n^2} \cdot \text{magnitude} \]
Since \(l = m = n = \frac{1}{\sqrt{3}}\), we have:
\[ \vec{n} = 2\sqrt{3} \left(\frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}\right) \]
Simplifying this gives:
\[ \vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k} \]
or
\[ \vec{n} = \pm 2\hat{i} \pm 2\hat{j} \pm 2\hat{k} \]

5. **Final Answer**:
Therefore, the vector \(\vec{n}\) can be expressed as:
\[ \vec{n} = \pm 2\left(\hat{i} + \hat{j} + \hat{k}\right) \]