Find the vector equation of plane through the points $(2, 1, - 1)$ and $(- 1, 3, 4)$ and perpendicular to the plane $x - 2 y + 4 z = 10$ .

\to r . (18^i + 17^j + 4^k) = 49

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\to r . (18^i + 15^j + 4^k) = 49

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\to r . (19^i + 17^j + 4^k) = 49

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\to r . (18^i + 17^j + 3^k) = 49

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Solution

The correct option is D $\to r . (18^i + 17^j + 4^k) = 49$ .
Consider the problem
Let,
the required plane passes through
$P (2, 1, - 1)$ and $Q (- 1, 3, 4)$
then
Position vector are
$a_{1} = 2 \to i + \to j - \to k$ and $a_{2} = - \to i + 3 \to j + 4 \to k$
then
$\Rightarrow P Q = (a_{2} - a_{1}) = - 3 \to i + 2 \to j + 5 \to k$
Let
$^i$ be the normal vector to the desired plane

$\to n = \to n_{i} \times - - \to P Q = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to i & \to j & \to k 1 & - 2 & 4 - 3 & 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\begin{matrix} \Rightarrow \to n = \to i (- 10 - 8) - \to j (5 + 12) + \to k (2 - 6) = - 18 \to i - 17 \to j - 4 \to k \end{matrix}$

$\begin{matrix} \Rightarrow \to r . \to n = \to a . \to n \Rightarrow \to r . (- 18 \to i - 17 \to j - 4 \to k) = (2 \to i + \to j - \to k) (- 18 \to i - 17 \to j - 4 \to k) \to r . (- 18 \to i - 17 \to j - 4 \to k) = - 36 - 17 + 4 \to r . (18 \to i + 17 \to j + 4 \to k) = 49 \end{matrix}$

Hence the option $A$ is the correct answer.

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\to r \cdot (2^i - 3^j + 4^k) = 1

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$\to r = 2^i + 4^j + 5^k + μ (3^i + 4^j + 5^k)$

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