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Question: The points \[A\left( -1,3,0 \right)\], \[B\left( 2,2,1 \right)\]and \[C\left( 1,1,3 \right)\] determ...

The points A(−1,3,0)A\left( -1,3,0 \right), B(2,2,1)B\left( 2,2,1 \right)and C(1,1,3)C\left( 1,1,3 \right) determine a plane. The distance of the plane ABCABC from the point D(5,7,8)D\left( 5,7,8 \right) is:
(a) 66\sqrt{66}
(b) 71\sqrt{71}
(c) 73\sqrt{73}
(d) 76\sqrt{76}

Explanation

Solution

We will first use the formula ∣x−x1y−y1z−z1 x2−x1y2−y1z2−z1 x3−x1y3−y1z3−z1 ∣=0\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0 to find the equation of the plane ABCABC passing through these points A, B, and C since we know from the question that the points A, B, and C determine a plane. Then we will find the distance of the point DD from this plane ABCABC.

Complete step-by-step solution:
We know that the equation of plane passing through any three points (x1,y1,z1),(x2,y2,z2),(x3,y3,z3)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right) is given by
∣x−x1y−y1z−z1 x2−x1y2−y1z2−z1 x3−x1y3−y1z3−z1 ∣=0\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0
Here, from the question, we get

& {{x}_{1}}=-1,{{x}_{2}}=2,{{x}_{3}}=1, \\\ & {{y}_{1}}=3,{{y}_{2}}=2,{{y}_{3}}=1, \\\ & {{z}_{1}}=0,{{z}_{2}}=1,{{z}_{3}}=3 \\\ \end{aligned}$$ Then the equation of plane $ABC$ passing through points A, B, and C is given by: $$\left| \begin{matrix} x-\left( -1 \right) & y-3 & z-0 \\\ 2-\left( -1 \right) & 2-3 & 1-0 \\\ 1-\left( -1 \right) & 1-3 & 3-0 \\\ \end{matrix} \right|=0$$ Solving it further, we get $$\begin{aligned} & \Rightarrow \left| \begin{matrix} x+1 & y-3 & z-0 \\\ 2+1 & 2-3 & 1-0 \\\ 1+1 & 1-3 & 3-0 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow \left| \begin{matrix} x+1 & y-3 & z \\\ 3 & -1 & 1 \\\ 2 & -2 & 3 \\\ \end{matrix} \right|=0 \\\ \end{aligned}$$ Expanding along the first row and solving further, we get $$\begin{aligned} & \Rightarrow \left( x+1 \right)\left\\{ \left( -1 \right)\cdot 3-1\cdot \left( -2 \right) \right\\}-\left( y-3 \right)\left\\{ 3\cdot 3-2\cdot 1 \right\\}+z\left\\{ 3\cdot \left( -2 \right)-2\left( -1 \right) \right\\}=0 \\\ & \Rightarrow \left( x+1 \right)\left( -3+2 \right)-\left( y-3 \right)\left( 7 \right)+z\left( -6+2 \right)=0 \\\ & \Rightarrow -\left( x+1 \right)-7\left( y-3 \right)-4z=0 \\\ & \Rightarrow -x-1-7y+21-4z=0 \\\ & \Rightarrow -x-7y-4z+20=0 \\\ & \Rightarrow x+7y+4z-20=0 \\\ \end{aligned}$$ We know that the distance of a point $\left( {{x}_{4}},{{y}_{4}},{{z}_{4}} \right)$ from a plane $ax+by+cz+d=0$ is given by $T=\left| \dfrac{a{{x}_{4}}+b{{y}_{4}}+c{{z}_{4}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$ Then the distance of the point D from the plane ABC is given by $\begin{aligned} & T=\left| \dfrac{5\cdot 1+7\cdot 7+8\cdot 4-20}{\sqrt{{{1}^{2}}+{{7}^{2}}+{{4}^{2}}}} \right|\\\ & =\left| \dfrac{5+49+32-20}{\sqrt{1+49+16}} \right| \\\ & =\left| \dfrac{66}{\sqrt{66}} \right| \\\ & =\sqrt{66} \end{aligned}$ **Hence, the correct option is (a).** **Note:** We should always be careful with the plus (+) and minus (-) sign while writing the formulas and calculating, because a single error can lead to change in the answer drastically. Also, practice by solving a few similar types of questions so that there is less confusion regarding the positions of $x,y,z,{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{y}_{1}},{{y}_{2}},{{y}_{3}},{{y}_{4}},{{z}_{1}},{{z}_{2}},{{z}_{3}},{{z}_{4}}$ in the above used formulas.