1. **Problem statement:** We are given points $A(1,1,0)$, $B(2,0,0)$, $C(1,3,-1)$, and $E(2,2,2)$. **a. Find the equation of plane $(P)$ determined by points $A$, $B$, and $C$.** Step 1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $$\overrightarrow{AB} = B - A = (2-1, 0-1, 0-0) = (1,-1,0)$$ $$\overrightarrow{AC} = C - A = (1-1, 3-1, -1-0) = (0,2,-1)$$ Step 2: Find the normal vector $\vec{n}$ to plane $(P)$ by the cross product: $$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 0 \\ 0 & 2 & -1 \end{vmatrix} = (-1)(-1) - 0(2),\; -(1(-1)-0(0)),\; 1(2) - (-1)(0) = (1,1,2)$$ Step 3: Write the equation of plane $(P)$ using normal vector $\vec{n} = (1,1,2)$ and point $A(1,1,0)$: $$1(x-1) + 1(y-1) + 2(z-0) = 0 \Rightarrow x + y + 2z - 2 = 0$$ **b. Show line $(AE)$ is perpendicular to plane $(P)$.** Step 1: Find vector $\overrightarrow{AE}$: $$\overrightarrow{AE} = E - A = (2-1, 2-1, 2-0) = (1,1,2)$$ Step 2: Compare $\overrightarrow{AE}$ with the normal vector $\vec{n}=(1,1,2)$: They are equal, so $\overrightarrow{AE}$ is parallel to $\vec{n}$, meaning line $(AE)$ is perpendicular to plane $(P)$. **c. Calculate area of triangle $ABC$ and volume of tetrahedron $EABC$.** Step 1: Area of triangle $ABC$ is half the magnitude of $\overrightarrow{AB} \times \overrightarrow{AC}$: $$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$$ Area: $$S = \frac{1}{2} \sqrt{6} = \frac{\sqrt{6}}{2}$$ Step 2: Volume of tetrahedron $EABC$ is \(\frac{1}{6}\) of the absolute value of the scalar triple product $|\overrightarrow{AE} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})|$. Calculate: $$\overrightarrow{AE} \cdot (\overrightarrow{AB} \times \overrightarrow{AC}) = (1,1,2) \cdot (1,1,2) = 1+1+4=6$$ Volume: $$V = \frac{|6|}{6} = 1$$ 2. **L is midpoint of segment [AB], (Q) plane passes through L and is parallel to lines (AE) and (BC).** **a. Write the equation of plane $(Q)$.** Step 1: Find midpoint $L$: $$L = \left(\frac{1+2}{2}, \frac{1+0}{2}, \frac{0+0}{2}\right) = (1.5, 0.5, 0)$$ Step 2: Find vector $\overrightarrow{BC}$: $$\overrightarrow{BC} = C - B = (1-2, 3-0, -1-0) = (-1,3,-1)$$ Step 3: Normal vector $\vec{n}_Q$ to plane $(Q)$ is perpendicular to both directions $(AE)$ and $(BC)$: $$\vec{n}_Q = \overrightarrow{AE} \times \overrightarrow{BC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ -1 & 3 & -1 \end{vmatrix}$$ Calculate: $$\vec{n}_Q = (1 \times (-1) - 2 \times 3, -(1 \times (-1) - 2 \times (-1)), 1 \times 3 - 1 \times (-1)) = (-1 -6, -( -1 + 2 ), 3 + 1) = (-7, -1, 4)$$ Step 4: Equation of plane $(Q)$ passing through $L(1.5,0.5,0)$ with normal vector $\vec{n}_Q$: $$-7(x-1.5) -1(y-0.5) + 4(z-0) = 0$$ Simplify: $$-7x + 10.5 - y + 0.5 + 4z = 0 \Rightarrow -7x - y + 4z + 11 = 0$$ **b. Show planes $(P)$ and $(Q)$ are perpendicular.** Step 1: Normal of plane $(P)$ is $\vec{n} = (1,1,2)$ and of $(Q)$ is $\vec{n}_Q = (-7,-1,4)$. Step 2: Compute dot product: $$\vec{n} \cdot \vec{n}_Q = 1 \times (-7) + 1 \times (-1) + 2 \times 4 = -7 -1 + 8 = 0$$ Since dot product is zero, planes are perpendicular. **c. Show line $(d)$, intersection of $(P)$ and $(Q)$, is parallel to $(BC)$.** Step 1: Direction vector of line $(d)$ is cross product of normals: $$\vec{d} = \vec{n} \times \vec{n}_Q = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ -7 & -1 & 4 \end{vmatrix}$$ Calculate: $$\vec{d} = (1 \times 4 - 2 \times (-1), -(1 \times 4 - 2 \times (-7)), 1 \times (-1) - 1 \times (-7)) = (4 + 2, -(4 + 14), -1 + 7) = (6, -18, 6)$$ Step 2: Simplify $\vec{d} = (6,-18,6) = 6(1,-3,1)$ which is parallel to $\overrightarrow{BC} = (-1,3,-1) = -1(1,-3,1)$. Therefore, line $(d)$ is parallel to line $(BC)$. **Final answers:** - Equation of plane $(P): x + y + 2z - 2 = 0$ - Line $(AE)$ is perpendicular to $(P)$ since $\overrightarrow{AE} = \vec{n}$ - Area of triangle $ABC = \frac{\sqrt{6}}{2}$ - Volume of tetrahedron $EABC = 1$ - Equation of plane $(Q): -7x - y + 4z + 11 = 0$ - $(P) \perp (Q)$ - Line of intersection $(d)$ of $(P)$ and $(Q)$ is parallel to $(BC)$.