Find the equation of the plane which contains the line of intersection of the planes:
`vec(r). (hat(i) - 2 hat(j) + 3 hat(k) ) - 4 = 0 ` and
`vec(r). ( - 2 hat(i) + hat(j) + hat(k)) + 5 = 0` and whose intercept on x-axis is equal to that of y-axis.

Answer

To find the equation of the plane that contains the line of intersection of the two given planes and has equal intercepts on the x-axis and y-axis, we can follow these steps:

### Step 1: Write the equations of the given planes in Cartesian form.

The first plane is given by:
\[ \vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) - 4 = 0 \]
This can be rewritten in Cartesian form as:
\[ x - 2y + 3z - 4 = 0 \quad \text{(Equation 1)} \]

The second plane is given by:
\[ \vec{r} \cdot (-2\hat{i} + \hat{j} + \hat{k}) + 5 = 0 \]
This can be rewritten in Cartesian form as:
\[ -2x + y + z + 5 = 0 \quad \text{(Equation 2)} \]

### Step 2: Find the line of intersection of the two planes.

The line of intersection of the two planes can be represented as a family of planes. We can express this as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x - 2y + 3z - 4) + \lambda (-2x + y + z + 5) = 0 \]
This simplifies to:
\[ (1 - 2\lambda)x + (-2 + \lambda)y + (3 + \lambda)z - (4 + 5\lambda) = 0 \]

### Step 3: Determine the intercepts condition.

We need to find the intercepts on the x-axis and y-axis. The intercepts can be found by setting the other variables to zero.

**X-intercept** (set \(y = 0\) and \(z = 0\)):
\[ (1 - 2\lambda)x - (4 + 5\lambda) = 0 \implies x = \frac{4 + 5\lambda}{1 - 2\lambda} \]

**Y-intercept** (set \(x = 0\) and \(z = 0\)):
\[ (-2 + \lambda)y - (4 + 5\lambda) = 0 \implies y = \frac{4 + 5\lambda}{-2 + \lambda} \]

### Step 4: Set the intercepts equal.

To satisfy the condition that the x-intercept equals the y-intercept:
\[ \frac{4 + 5\lambda}{1 - 2\lambda} = \frac{4 + 5\lambda}{-2 + \lambda} \]

Cross-multiplying gives:
\[ (4 + 5\lambda)(-2 + \lambda) = (4 + 5\lambda)(1 - 2\lambda) \]
Assuming \(4 + 5\lambda \neq 0\), we can divide both sides by \(4 + 5\lambda\):
\[ -2 + \lambda = 1 - 2\lambda \]
Solving for \(\lambda\):
\[ 3\lambda = 3 \implies \lambda = 1 \]

### Step 5: Substitute \(\lambda\) back into the plane equation.

Now substituting \(\lambda = 1\) back into our family of planes:
\[ (1 - 2 \cdot 1)x + (-2 + 1)y + (3 + 1)z - (4 + 5 \cdot 1) = 0 \]
This simplifies to:
\[ -1x - 1y + 4z - 9 = 0 \]
Rearranging gives:
\[ x + y - 4z + 9 = 0 \]

### Final Equation of the Plane

Thus, the equation of the required plane is:
\[ x + y - 4z + 9 = 0 \]

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