vec a=\2λ,1,1-λ\ , vec b=\-1,3,0\ , vec c=\5,-1,8\ . 1^°…
$\vec a=\{2\lambda,1,1-\lambda\}$, $\vec b=\{-1,3,0\}$, $\vec c=\{5,-1,8\}$. $1^\circ$ Find $\lambda$ so that $\vec a$ makes equal angles with $\vec b$ and $\vec c$. $2^\circ$ For that $\lambda$, find the inclination of $\vec c$ to the plane of $\vec a,\vec b$. $3^\circ$ Find the volume and one height of the parallelepiped on $\vec a,\vec b,\vec c$.
Explanation of Method
Equal angles, the inclination of c to a plane and a parallelepiped volume.
Step 1: $1^\circ$ $\lambda$ (as in Problem 71)
\[ \frac{-2\lambda+3}{\sqrt{10}}=\frac{2\lambda+7}{3\sqrt{10}}\ \Longrightarrow\ \lambda=\frac14,\qquad \vec a=\left\{\tfrac12,\ 1,\ \tfrac34\right\}. \]
Step 2: Normal of the plane and triple product
\[ \vec a\times\vec b=\left\{-\tfrac94,\ -\tfrac34,\ \tfrac52\right\}=\tfrac14\{-9,-3,10\},\qquad |\vec a\times\vec b|=\frac{\sqrt{190}}{4}, \]
\[ [\vec a,\vec b,\vec c]=\tfrac14(-45+3+80)=\frac{19}{2},\qquad |\vec c|=3\sqrt{10}. \]
Step 3: $2^\circ$ Inclination
The angle $\psi$ between $\vec c$ and the plane is the complement of the angle between $\vec c$ and the normal: \[\sin\psi=\frac{|[\vec a,\vec b,\vec c]|}{|\vec a\times\vec b|\,|\vec c|}=\frac{19/2}{\frac{\sqrt{190}}{4}\cdot3\sqrt{10}}=\frac{38}{30\sqrt{19}}=\frac{\sqrt{19}}{15}\approx0.2906.\]
\[\psi=\arcsin\frac{\sqrt{19}}{15}\approx16.9^\circ\]
Step 4: $3^\circ$ Volume and height
\[V=\frac{19}{2},\qquad h_c=\frac{V}{|\vec a\times\vec b|}=\frac{38}{\sqrt{190}}=\frac{\sqrt{190}}{5}\approx2.76\]
Check: $h_c=|\vec c|\sin\psi=3\sqrt{10}\cdot\dfrac{\sqrt{19}}{15}=\dfrac{\sqrt{190}}{5}$ $\checkmark$
Result
\[\boxed{\,\psi=\arcsin\frac{\sqrt{19}}{15}\approx16.9^\circ\,}\]
\[\boxed{\,V=\frac{19}{2},\qquad h_c=\frac{V}{|\vec a\times\vec b|}=\frac{38}{\sqrt{190}}=\frac{\sqrt{190}}{5}\approx2.76\,}\]
Remark
Correction. $\arcsin(0.2906)=16.89^\circ$, not $16.3^\circ$.