Find the point , where the line joining the points (1,3,4) and (-3,5,2) intersects the plane `vec(r) . (2 hati + hatj + hatk) + 3 = 0 . `
Is the point equidistant from the given points ?

Answer

To find the point where the line joining the points \( A(1, 3, 4) \) and \( B(-3, 5, 2) \) intersects the plane given by the equation \( \vec{r} \cdot (2 \hat{i} + \hat{j} + \hat{k}) + 3 = 0 \), we can follow these steps:

### Step 1: Find the direction ratios of the line
The direction ratios of the line joining points \( A \) and \( B \) can be determined by subtracting the coordinates of \( A \) from those of \( B \):
\[ \text{Direction ratios} = B - A = (-3 - 1, 5 - 3, 2 - 4) = (-4, 2, -2) \]

### Step 2: Write the parametric equations of the line
Using point \( A(1, 3, 4) \) as a point on the line and the direction ratios we found, we can write the parametric equations:
\[ x = 1 - 4\lambda, \quad y = 3 + 2\lambda, \quad z = 4 - 2\lambda \]

### Step 3: Substitute the parametric equations into the plane equation
The equation of the plane is given as:
\[ \vec{r} \cdot (2 \hat{i} + \hat{j} + \hat{k}) + 3 = 0 \]
This can be expanded to:
\[ 2x + y + z + 3 = 0 \]
Substituting the parametric equations into the plane equation:
\[ 2(1 - 4\lambda) + (3 + 2\lambda) + (4 - 2\lambda) + 3 = 0 \]
Simplifying this:
\[ 2 - 8\lambda + 3 + 2\lambda + 4 - 2\lambda + 3 = 0 \]
Combine like terms:
\[ 12 - 8\lambda = 0 \]
Thus, solving for \( \lambda \):
\[ 8\lambda = 12 \implies \lambda = \frac{12}{8} = \frac{3}{2} \]

### Step 4: Find the coordinates of the intersection point
Now, substitute \( \lambda = \frac{3}{2} \) back into the parametric equations to find the coordinates of point \( P \):
\[ x = 1 - 4\left(\frac{3}{2}\right) = 1 - 6 = -5 \]
\[ y = 3 + 2\left(\frac{3}{2}\right) = 3 + 3 = 6 \]
\[ z = 4 - 2\left(\frac{3}{2}\right) = 4 - 3 = 1 \]
Thus, the coordinates of point \( P \) are \( (-5, 6, 1) \).

### Step 5: Check if point \( P \) is equidistant from points \( A \) and \( B \)
To check if point \( P \) is equidistant from points \( A \) and \( B \), we calculate the distances \( d_1 \) (from \( A \) to \( P \)) and \( d_2 \) (from \( B \) to \( P \)).

**Distance \( d_1 \) from \( A(1, 3, 4) \) to \( P(-5, 6, 1) \)**:
\[ d_1 = \sqrt{(-5 - 1)^2 + (6 - 3)^2 + (1 - 4)^2} = \sqrt{(-6)^2 + (3)^2 + (-3)^2} = \sqrt{36 + 9 + 9} = \sqrt{54} = 3\sqrt{6} \]

**Distance \( d_2 \) from \( B(-3, 5, 2) \) to \( P(-5, 6, 1) \)**:
\[ d_2 = \sqrt{(-5 + 3)^2 + (6 - 5)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Since \( d_1 \neq d_2 \), point \( P \) is not equidistant from points \( A \) and \( B \).

### Final Answer
The point of intersection \( P \) is \( (-5, 6, 1) \) and it is not equidistant from points \( A \) and \( B \).