The triangle ABC is defined by the vertices $\mathrm{A} = (0, 7, 10), \mathrm{B} = (-1, 6, 6)$ and $\mathrm{C} = (-4,9, 6)$ Let D be the foot of the altitude from B to the side $\mathrm{AC}$. Then $\overline{\mathrm{BD}}$ is

$A=(0,7,10),B=(-1,6,6) and C=(-4,9,6)$

$P.V of AB is (-1,-1,-4)$

$⟹\left|AB\right|=\sqrt{\left(\right(-1{)}^2+(-1{)}^2+(-4{)}^2)}=\sqrt{18}$

$P.V. of BC is (-3,3,0)$

$⟹\left|BC\right|=\left(\left(-3^2\right)+3^2+0\right)=\sqrt{18P}.$

$P.V of AC is (-4,2,-4)$

$⟹\left|AC\right|=\sqrt{\left(\right(-4{)}^2+2^2+(-4{)}^2)}=6$

$Now, D is the midpoint of AC$

$⟹D=\left(\frac{(0-4)}{2},\frac{(7+9)}{2},\frac{(10+6)}{2}\right)=(-2,8,8)$

$⟹\left(\overline{BD}\right)=-\overline{i}+2\overline{j}+2\overline{k}$