Let the equation of the plane P containing the line x+10=8-y2=z be ax+by+3z=2(a+b), and th…

Question

Let the equation of the plane $P$ containing the line $x+10=\dfrac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$, and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to:

Answer: 355

Accepted Answer

The line $\dfrac{x+10}{1}=\dfrac{y-8}{-2}=\dfrac{z}{1}$ passes through $(-10,8,0)$ with direction ratios $(1,-2,1)$.

Since the plane contains this line, the point satisfies the plane equation, giving $b=2a$, and the normal $(a,b,3)$ is perpendicular to the direction:
$$a-2b+3=0 \implies a-2(2a)+3=0 \implies a=1,\ b=2.$$
$$\text{Plane: } x+2y+3z=6.$$
$$c=\frac{|1+54+21-6|}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5\sqrt{14}.$$
$a^2+b^2+c^2=1+4+350=355.$

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