Let the line L pass through the point (0, 1, 2) , intersect the line x−12=y−23=z−34 and be parallel to the plane 2x + y - 3z = 4 . Then the distance of the point P(1, -9, 2) from the line L is

x−12=y−23=z−34=rAny point on the line isDr’s of AB↔ are(2r + 1, 3r + 1, 4r + 1)∵AB→∥le to the plane 2x + y - 3z = 4⇒al+bm+cn=0⇒2(2r+1)+1(3r+1)−3(4r+1)=0⇒r=0B(1, 2, 3)L′=x−01=y−11=z−21=t⇒Q(t,t+1,t+2);P(1,−9,2)Dr’s of PQ↔ = (t - 1)1 + (t + 10)1 + t = 0PQ¯=(−4,7,−3)t=−3|PQ¯|=16+49+9=74

Let the line L pass through the point (0, 1, 2) , intersect the line x−12=y−23=z−34 and be parallel to the plane 2x + y - 3z = 4 . Then the distance of the point P(1, -9, 2) from the line L is

x−12=y−23=z−34=rAny point on the line isDr’s of AB↔ are(2r + 1, 3r + 1, 4r + 1)∵AB→∥le to the plane 2x + y - 3z = 4⇒al+bm+cn=0⇒2(2r+1)+1(3r+1)−3(4r+1)=0⇒r=0B(1, 2, 3)L′=x−01=y−11=z−21=t⇒Q(t,t+1,t+2);P(1,−9,2)Dr’s of PQ↔ = (t - 1)1 + (t + 10)1 + t = 0PQ¯=(−4,7,−3)t=−3|PQ¯|=16+49+9=74

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Let the line L pass through the point (0, 1, 2) , intersect the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and be parallel to the plane
2x + y - 3z = 4 . Then the distance of the point P(1, -9, 2) from the line L is

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$\sqrt{74}$

$\sqrt{54}$

$\sqrt{69}$

answer is D.

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Detailed Solution

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = r$
Any point on the line is
Question Image
Dr’s of $\overset{\leftrightarrow}{AB}$ are
(2r + 1, 3r + 1, 4r + 1)
$∵ \vec{AB} ∥^{le}$ to the plane 2x + y - 3z = 4
$\begin{array}{l} \Rightarrow al + bm + cn = 0 \\ \Rightarrow 2 (2 r + 1) + 1 (3 r + 1) - 3 (4 r + 1) = 0 \\ \Rightarrow r = 0 \end{array}$
B(1, 2, 3)
$\begin{array}{l} L^{'} = \frac{x - 0}{1} = \frac{y - 1}{1} = \frac{z - 2}{1} = t \\ \Rightarrow Q (t, t + 1, t + 2); P (1, - 9, 2) \end{array}$
Dr’s of $\overset{\leftrightarrow}{PQ}$ = (t - 1)1 + (t + 10)1 + t = 0
$\begin{array}{l} \bar{PQ} = (- 4, 7, - 3) t = - 3 \\ | \bar{PQ} | = \sqrt{16 + 49 + 9} = \sqrt{74} \end{array}$