The point of intersection of the line 3x−1=4y+2=−2z−3 and the plane 2x−y+3z−1=0, is

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The point of intersection of the line {x-1}{3}={y+2}{4}={z-3}{

Question

The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{- 2}$ and the plane $2 x - y + 3 z - 1 = 0$ , is

(- 10, 10, 3)

(10, 10, - 3)

(10, - 10, 3)

(10, - 10, - 3)

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Elijah Discussed

The point of intersection of the line

\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{- 2}

and the plane

2 x - y + 3 z - 1 = 0

, is

5 mins ago

Discuss this question LIVE

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Question Text

The point of intersection of the line

\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{- 2}

and the plane

2 x - y + 3 z - 1 = 0

, is