Find the point R, Where the line joining P (1,3,4) and Q (-3,5,2) cuts the plane `vec(r). ( 2 hati - hatj + hatk) + 3 = 0`.
is `|vec(PR)| = |vec(QR)|` ?

Answer

To find the point \( R \) where the line joining points \( P(1, 3, 4) \) and \( Q(-3, 5, 2) \) cuts the plane defined by the equation \( \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 3 = 0 \) and satisfies the condition \( |\vec{PR}| = |\vec{QR}| \), we can follow these steps:

### Step 1: Parametrize the Line Segment PQ
The line joining points \( P \) and \( Q \) can be expressed in parametric form. Let \( R \) be a point on the line segment \( PQ \) such that:

\[ R = P + t(Q - P) \]

Where \( t \) is a parameter that varies from 0 to 1. The coordinates of \( R \) can be calculated as:

\[ R = (1, 3, 4) + t((-3, 5, 2) - (1, 3, 4)) \]
\[ R = (1, 3, 4) + t(-4, 2, -2) \]
\[ R = (1 - 4t, 3 + 2t, 4 - 2t) \]

### Step 2: Substitute R into the Plane Equation
We substitute the coordinates of \( R \) into the plane equation:

\[ \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 3 = 0 \]

Substituting \( R = (1 - 4t, 3 + 2t, 4 - 2t) \):

\[ 2(1 - 4t) - (3 + 2t) + (4 - 2t) + 3 = 0 \]

Expanding this gives:

\[ 2 - 8t - 3 - 2t + 4 - 2t + 3 = 0 \]
\[ (2 - 3 + 4 + 3) + (-8t - 2t - 2t) = 0 \]
\[ 6 - 12t = 0 \]

### Step 3: Solve for t
Now, we solve for \( t \):

\[ 12t = 6 \implies t = \frac{1}{2} \]

### Step 4: Find the Coordinates of R
Now, we substitute \( t = \frac{1}{2} \) back into the equation for \( R \):

\[ R = (1 - 4 \cdot \frac{1}{2}, 3 + 2 \cdot \frac{1}{2}, 4 - 2 \cdot \frac{1}{2}) \]
\[ R = (1 - 2, 3 + 1, 4 - 1) \]
\[ R = (-1, 4, 3) \]

### Step 5: Verify the Lengths PR and QR
Next, we need to verify that \( |\vec{PR}| = |\vec{QR}| \).

1. Calculate \( \vec{PR} \):
\[ \vec{PR} = R - P = (-1 - 1, 4 - 3, 3 - 4) = (-2, 1, -1) \]
\[ |\vec{PR}| = \sqrt{(-2)^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

2. Calculate \( \vec{QR} \):
\[ \vec{QR} = R - Q = (-1 + 3, 4 - 5, 3 - 2) = (2, -1, 1) \]
\[ |\vec{QR}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Since \( |\vec{PR}| = |\vec{QR}| \), the condition is satisfied.

### Final Answer
The point \( R \) where the line joining \( P \) and \( Q \) cuts the plane is:

\[ \boxed{(-1, 4, 3)} \]