Find the equation of the plane passing through the point \( (2,-3,2) \) and the line of intersection of the two planes: \( 6 x+4 y+3 z+5=0,2 x+y+z=2 \) \( \begin{array}{ll}\text { a) }(16,7,8) \cdot \vec{r}=1 & \text { b) }(16,7,8) \cdot \vec{r}=27 \\ \text { c) }(16,7,8) \cdot \vec{r}=54 & \text { d) }(16,-7,8) \cdot \vec{r}=27\end{array} \)

To find the equation of the plane passing through the point \( (2, -3, 2) \) and the line of intersection of the two given planes, we start by determining the normal vector of the desired plane. First, we find the normal vectors of the given planes: 1. For the plane \( 6x + 4y + 3z + 5 = 0 \), the normal vector \( \vec{n_1} = (6, 4, 3) \). 2. For the plane \( 2x + y + z = 2 \), the normal vector \( \vec{n_2} = (2, 1, 1) \). Next, we can find the direction vector of the line of intersection by computing the cross product of \( \vec{n_1} \) and \( \vec{n_2} \): \[ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 3 \\ 2 & 1 & 1 \\ \end{vmatrix} = \hat{i}(4·1 - 3·1) - \hat{j}(6·1 - 3·2) + \hat{k}(6·1 - 4·2) = \hat{i}(4 - 3) - \hat{j}(6 - 6) + \hat{k}(6 - 8) = (1, 0, -2). \] Now we have a direction vector \( \vec{d} = (1, 0, -2) \). The normal vector of the desired plane can also be obtained by using the linear combination of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \). Let's take \( \vec{n} = \vec{n_1} + k \cdot \vec{n_2} \) for some scalar \( k \). Instead, for simplicity, we can choose \( k = -1 \). Thus, \[ \vec{n} = (6, 4, 3) + (-1)(2, 1, 1) = (4, 3, 2). \] With the normal vector \( \vec{n} = (4, 3, 2) \), we can use the point-normal form of the equation of the plane: \[ 4(x - 2) + 3(y + 3) + 2(z - 2) = 0. \] Expanding this yields: \[ 4x - 8 + 3y + 9 + 2z - 4 = 0 \implies 4x + 3y + 2z - 3 = 0. \] Rearranging gives the equation of the plane: \[ 4x + 3y + 2z = 3. \] Now to compare this to the given options, we can check if any of the choices can represent this plane. Rearranging the given options, we find that: - Checking option (b): \( (16, 7, 8) \cdot (x, y, z) = 27 \) --> \( 16x + 7y + 8z = 27 \) does not match, - Checking option (c): \( (16, 7, 8) \cdot (x, y, z) = 54 \) --> \( 16x + 7y + 8z = 54 \) does not match, - Checking option (d): \( (16, -7, 8) \cdot (x, y, z) = 27 \) --> \( 16x - 7y + 8z = 27 \) does not match, - Checking option (a): \( (16, 7, 8) \cdot (x, y, z) = 1\) --> \( 16x + 7y + 8z = 1 \) does not match, As we find that none match, we would therefore refine our calculations. Let's try another linear combination for \( k = 2 \): \[ \vec{n} = (6, 4, 3) - 2(2, 1, 1) = (2, 2, 1). \] We therefore write: \[ 2(x - 2) + 2(y + 3) + 1(z - 2) =