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(B) The equation of a plane passing through a point $(x_1, y_1, z_1)$ with a normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x

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$x - 2y + 3z = 14$

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(B) The equation of a plane passing through a point $(x_1, y_1, z_1)$ with a normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Given point is $(1, -2, 3)$ and normal vector is $\vec{n} = -\hat{i} + 2\hat{j} - 3\hat{k}$. Substituting the values, we get: $-1(x - 1) + 2(y - (-2)) - 3(z - 3) = 0$ $-1(x - 1) + 2(y + 2) - 3(z - 3) = 0$ $-x + 1 + 2y + 4 - 3z + 9 = 0$ $-x + 2y - 3z + 14 = 0$ $x - 2y + 3z = 14$.

The Cartesian equation of the plane passing through the point $(1, -2, 3)$ and perpendicular to the vector $-\hat{i} + 2\hat{j} - 3\hat{k}$ is:

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