Find the equation of the plane passing through the point A(2, 1, 3), B(-1, 2, 4) and C(0, 2, 1). Also determine its point of intersection with the line $r = i - j + k + t (2 i + k) .$

(7, - 1, 4) .

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(7, 1, 4) .

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(7, - 1, - 4) .

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(- 7, 1, - 4) .

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Solution

The correct option is A $(7, - 1, 4) .$
If $n$ be the normal of the plane $A B C,$ then
$n = \to A B \times \to A C = - 3 i - 8 j - k$
Plane is $(r - a) \cdot n = 0$ $\Rightarrow r \cdot n = a \cdot n$
$\Rightarrow r \cdot (3 i + 8 j + k) = 17$ ...(1)
The given line is $r = (1 + 2 t) i + (- 1) j + (1 + t) k = 0$ ...(2)
For point of intersection we have from (1) and (2)
$3 (1 + 2 t) - 8 + (1 + t) = 17 ∴ 7 t = 21$ or t = 3
Substitute for $t$ in (2) the point of intersection is given by $r = 7 i - j + 4 k .$
$∴$ Its co-ordinates are $(7, - 1, 4) .$

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Find the equation of the plane passing through the point A(2, 1, 3), B(-1, 2, 4) and C(0, 2, 1). Also determine its point of intersection with the line r=i−j+k+t(2i+k).

Find the equation of the plane passing through the point A(2, 1, 3), B(-1, 2, 4) and C(0, 2, 1). Also determine its point of intersection with the line $r = i - j + k + t (2 i + k) .$