Step 1: Recall what makes a raster/image operation linear.
An image processing operation is linear if the output at every pixel can be expressed as a weighted sum (convolution) of input pixel values, i.e. it satisfies superposition: \(T(a f_1 + b f_2) = a\,T(f_1) + b\,T(f_2)\). Spatial filters implemented as a fixed convolution kernel applied uniformly over the image are linear operations.
Step 2: Test each option against this definition.
(B) An averaging (mean/box) filter replaces each pixel by the average of its neighbourhood, a convolution with a constant-weight kernel, so it is linear. (C) Convolution with a Gaussian kernel is, by definition, a convolution with fixed weights, so it is linear. (D) Laplacian edge detection approximates the second spatial derivative using a fixed convolution mask, e.g. \(\begin{bmatrix}0 & 1 & 0\\ 1 & -4 & 1\\ 0 & 1 & 0\end{bmatrix}\); being a convolution with constant coefficients, it is also linear.
Step 3: Examine histogram equalization.
Histogram equalization is a point (radiometric) operation that remaps each grey level using a transformation derived from the image's own cumulative distribution function (CDF): \[ s = T(r) = (L-1)\sum_{j=0}^{r}\frac{n_j}{n} \] This mapping depends on the image's own histogram and is generally a non-linear, monotonically increasing step function; it does not satisfy superposition, so it is non-linear.
Step 4: Conclude.
Since averaging, Gaussian convolution and Laplacian edge detection are all linear (fixed-kernel convolutions), while histogram equalization is a non-linear, data-dependent remapping, the operation that is NOT linear is histogram equalization. \[ \boxed{\text{Histogram equalization}} \]