Answer
$$6$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1, - 1,1} \right)} \frac{{xz + 5x + yz + 5y}}{{x + y}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{xz + 5x + yz + 5y}}{{x + y}} = \frac{{\left( 1 \right)\left( 1 \right) + 5\left( 1 \right) + \left( { - 1} \right)\left( 1 \right) + 5\left( { - 1} \right)}}{{1 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 + 5 - 1 - 5}}{{1 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{0}{0} \cr
& {\text{Factoring the numerator by grouping terms}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\left( {xz + 5x} \right) + \left( {yz + 5y} \right)}}{{x + y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{x\left( {z + 5} \right) + y\left( {z + 5} \right)}}{{x + y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \frac{{\left( {z + 5} \right)\left( {x + y} \right)}}{{x + y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \left( {z + 5} \right) \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {1,1,1} \right)} \left( {z + 5} \right) = 1 + 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 \cr} $$
