Answer
$238$
Work Step by Step
Recall the dot product property for n-dimensional vectors, which states that if $\vec{u}=(u_1,u_2.....u_n)$ and $\vec{v}=(v_1,v_2.....v_n)$, then their dot product is:
$\vec{u}\cdot \vec{v}=u_1v_1+u_2v_2+........u_nv_n$
We are given that $\vec{a}=2j+k = \lt 0,2,1\gt; \vec{y}=4i-7j=\lt4,-7,0\gt; \vec{c}=i+6j =\lt 1,6,0\gt$ and $\vec{z}=i-3j-k=\lt1,-3,-1 \gt$
Our aim is to find the dot product $(\vec{z} \cdot \vec{c}) (\vec{y} \cdot \vec{a})$.
Now, $(\vec{z}\cdot \vec{c})=\lt 1,-3,-1 \gt \cdot \lt 1,6,0\gt=(1)(1)+(-3)(6)+(-1)(0) = -17$
$(\vec{y} \cdot \vec{a})= \lt 4,-7,0 \gt \cdot \lt 0,2,1\gt \\=(4)(0)+(-7)(2)+(0)(1) \\=-14$
Therefore, we have: $(\vec{z} \cdot \vec{c}) (\vec{y} \cdot \vec{a})=(-17)(-14)=238$
