Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{{x^2} + xy - xz - yz}}{{x - z}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{{x^2} + xy - xz - yz}}{{x - z}} = \frac{{{{\left( 1 \right)}^2} + \left( 1 \right)\left( 1 \right) - \left( 1 \right)\left( 1 \right) - \left( 1 \right)\left( 1 \right)}}{{1 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 + 1 - 1 - 1}}{{1 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{0}{0} \cr
& {\text{Factoring the numerator by grouping terms}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\left( {{x^2} + xy} \right) - \left( {xz + yz} \right)}}{{x - z}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{x\left( {x + y} \right) - z\left( {x + y} \right)}}{{x - z}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\left( {x + y} \right)\left( {x - z} \right)}}{{x - z}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \left( {x + y} \right) \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \left( {x + y} \right) = 1 + 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cr} $$
