To solve the problem, we need to find the value of \( k \) such that the line given by the equation
\[
\frac{x}{k} = \frac{y}{2} = \frac{z}{-12}
\]
forms an isosceles triangle with the planes given by the equations
\[
2x + y + 3z - 1 = 0
\]
and
\[
x + 2y - 3z - 1 = 0.
\]
### Step 1: Write the equations of the planes and line
The line can be expressed in parametric form as:
\[
x = kt, \quad y = 2t, \quad z = -12t
\]
where \( t \) is a parameter.
### Step 2: Find the equations of the bisectors of the planes
The bisector of two planes can be found using the formula:
\[
\frac{A_1 x + B_1 y + C_1 z + D_1}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \pm \frac{A_2 x + B_2 y + C_2 z + D_2}{\sqrt{A_2^2 + B_2^2 + C_2^2}}
\]
For the first plane \( 2x + y + 3z - 1 = 0 \):
- \( A_1 = 2, B_1 = 1, C_1 = 3, D_1 = -1 \)
For the second plane \( x + 2y - 3z - 1 = 0 \):
- \( A_2 = 1, B_2 = 2, C_2 = -3, D_2 = -1 \)
### Step 3: Set up the bisector equation
The equation of the bisector becomes:
\[
\frac{2x + y + 3z - 1}{\sqrt{2^2 + 1^2 + 3^2}} = \pm \frac{x + 2y - 3z - 1}{\sqrt{1^2 + 2^2 + (-3)^2}}
\]
Calculating the denominators:
\[
\sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}, \quad \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}
\]
Thus, the equation simplifies to:
\[
2x + y + 3z - 1 = \pm (x + 2y - 3z - 1)
\]
### Step 4: Solve for the two cases
**Case 1: Positive sign**
\[
2x + y + 3z - 1 = x + 2y - 3z + 1
\]
Rearranging gives:
\[
x - y + 6z - 2 = 0 \quad \text{(Equation 1)}
\]
**Case 2: Negative sign**
\[
2x + y + 3z - 1 = - (x + 2y - 3z - 1)
\]
Rearranging gives:
\[
3x + 3y - 2 = 0 \quad \text{(Equation 2)}
\]
### Step 5: Determine which bisector is valid
The line must be parallel to one of these bisectors. The line's direction ratios are \( k, 2, -12 \).
For Equation 1 \( (x - y + 6z = 2) \):
The coefficients are \( 1, -1, 6 \).
For Equation 2 \( (3x + 3y - 2 = 0) \):
The coefficients are \( 3, 3, 0 \).
### Step 6: Compare coefficients
Since the line has no \( z \) term in Equation 2, it cannot be parallel to it. Thus, we check Equation 1:
\[
\frac{k}{1} = \frac{2}{-1} = \frac{-12}{6}
\]
From \( \frac{2}{-1} = -2 \), we have \( k = -2 \).
### Final Answer
Thus, the value of \( k \) is
\[
\boxed{-2}.
\]
