The plane `2x-2y+z+12=0` touches the surface `x^2+y^2+z^2-2x-4y+2z-3=0` only at the point `(-1, lambda, -2)`. The value of `lambda` must be ___

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the point \((-1, \lambda, -2)\) lies on the plane defined by the equation \(2x - 2y + z + 12 = 0\) and also satisfies the condition that the plane touches the surface defined by the equation \(x^2 + y^2 + z^2 - 2x - 4y + 2z - 3 = 0\).

### Step 1: Verify that the point lies on the plane
We start by substituting the coordinates of the point \((-1, \lambda, -2)\) into the plane equation.

The plane equation is:
\[ 2x - 2y + z + 12 = 0 \]
Substituting \(x = -1\), \(y = \lambda\), and \(z = -2\):
\[ 2(-1) - 2(\lambda) + (-2) + 12 = 0 \]
This simplifies to:
\[ -2 - 2\lambda - 2 + 12 = 0 \]
Combining like terms:
\[ 8 - 2\lambda = 0 \]
Rearranging gives:
\[ 2\lambda = 8 \]
Thus:
\[ \lambda = \frac{8}{2} = 4 \]

### Step 2: Conclusion
The value of \( \lambda \) must be \( 4 \).

### Final Answer
The value of \( \lambda \) is \( 4 \).

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