Let the equation of the plane P containing the line x+10=(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 __________.

The line (x+10 / 1)=(y-8 / -2)=(z / 1) have a point (-10,8,0) with d. r. (1,-2,1) Because the plane a x+b y+3 z=2(a+b) => b=2 a \ dot product of d.r.'s is zero Therefore a-2 b+3=0 Therefore a=1 \ b=2 Distance from (1,27,7) is c=frac1+54+21-6sqrt(14)=frac70sqrt(14)=5 sqrt(14) Therefore a^2+ b^2+ c^2=1+4+350 =355

Let the equation of the plane P containing the line x10... | null (Mathematics)

The line

\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}

have a point

(-10,8,0)

with d. r.

(1,-2,1)

\because

the plane

a x+b y+3 z=2(a+b)

\Rightarrow \mathrm{b}=2 \mathrm{a}

\&

dot product of d.r.'s is zero

\therefore \mathrm{a}-2 \mathrm{~b}+3=0

\therefore \mathrm{a}=1 \& \mathrm{~b}=2

Distance from

(1,27,7)

c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}

\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1+4+350

=355

Explanation

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