Let P be the plane containing the straight line x−39=y+4−1=z−7−5and perpendicular to the plane containing the straight lines x2=y3=z5 and x3=y7=z8. If d is the distance of P from the point (2, –5, 11), then d2 is equal to :

a(x−3)+b(y+4)+c(z−7)=0P: 9a−b−5c=0drs of line perpendicular to the given lines=ijk235378 =(-11,-1,5) ⇒−11a−b+5c=0After solving DR’s (a,b,c) ∝(1,−1,2)Equation of planex−y+2z=21Perpendicular distance from (2,-5,11)=2+5+22-211+1+4d=86d2=323

Let P be the plane containing the straight line x−39=y+4−1=z−7−5and perpendicular to the plane containing the straight lines x2=y3=z5 and x3=y7=z8. If d is the distance of P from the point (2, –5, 11), then d2 is equal to :

a(x−3)+b(y+4)+c(z−7)=0P: 9a−b−5c=0drs of line perpendicular to the given lines=ijk235378 =(-11,-1,5) ⇒−11a−b+5c=0After solving DR’s (a,b,c) ∝(1,−1,2)Equation of planex−y+2z=21Perpendicular distance from (2,-5,11)=2+5+22-211+1+4d=86d2=323

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Let P be the plane containing the straight line $\frac{x - 3}{9} = \frac{y + 4}{- 1} = \frac{z - 7}{- 5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2} = \frac{y}{3} = \frac{z}{5}$ and $\frac{x}{3} = \frac{y}{7} = \frac{z}{8}$ . If d is the distance of P from the point (2, –5, 11), then d² is equal to :

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$96$

$\frac{32}{3}$

$\frac{147}{2}$

$54$

answer is C.

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Detailed Solution

$\begin{array}{l} a (x - 3) + b (y + 4) + c (z - 7) = 0 \\ P : 9 a - b - 5 c = 0 \\ d r s o f l i n e p e r p e n d i c u l a r t o t h e g i v e n l i n e s = |\begin{matrix} i & j & k \\ 2 & 3 & 5 \\ 3 & 7 & 8 \end{matrix}| \end{array} = (- 11, - 1, 5) \Rightarrow - 11 a - b + 5 c = 0$

After solving DR’s (a,b,c) $\propto (1, - 1, 2)$

Equation of plane

$\begin{array}{l} x - y + 2 z = 21 \\ P e r p e n d i c u l a r d i s \tan c e f r o m (2, - 5, 11) \\ = \frac{|2 + 5 + 22 - 21|}{\sqrt{1 + 1 + 4}} \\ d = \frac{8}{\sqrt{6}} \\ d^{2} = \frac{32}{3} \end{array}$