University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{9}{\sqrt {41}}$
The formula to calculate the distance for two vectors is given by: $d=\dfrac{|u \cdot v|}{|v|}$ Thus, we have $u \cdot v=7(1)+(-2)(2)+1(6)=9$ and $|u \times v|=|9|=9$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{9}{ \sqrt {(1)^2+(2)^2+(6)^2}}=\dfrac{9}{\sqrt {41}}$