The equation of a plane containing the point | vedclass

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(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$.Since the plane is perpendicular to the plan

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$\bar{r} \cdot (5\hat{i} - 4\hat{j} - \hat{k}) = 7$

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(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$.Since the plane is perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (2, 3, -2)$ and $\vec{n_2} = (1, 2, -3)$.Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -2 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - (-4)) - \hat{j}(-6 - (-2)) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + \hat{k}$.So,$a = -5, b = 4, c = 1$.Substituting these values into the plane equation: $-5(x - 1) + 4(y + 1) + 1(z - 2) = 0$.$-5x + 5 + 4y + 4 + z - 2 = 0 \Rightarrow -5x + 4y + z + 7 = 0 \Rightarrow 5x - 4y - z = 7$.In vector form,this is $\bar{r} \cdot (5\hat{i} - 4\hat{j} - \hat{k}) = 7$.

The equation of a plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is

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