The equation of plane which is passing through \((2,1,3)\) and having equal \(X\) and Y-intercept and \(Z\) -intercept 14 is (A) \(11 \mathrm{x}-11 \mathrm{y}+3 \mathrm{z}=42\) (B) \(11 \mathrm{x}+11 \mathrm{y}+3 z=42\) (C) \(11 \mathrm{x}+11 \mathrm{y}-3 \mathrm{z}=42\) (D) \(11 \mathrm{x}+11 \mathrm{y}+3 \mathrm{z}+42=0\)

We are given the plane's intercepts, which are the points at which the plane intersects the axes. Based on the information, the X and Y intercepts are equal, meaning the points are \((A,A,0)\) and \((0,A,0)\). Since the Z-intercept is given as 14, the point \((0,0,14)\). As we know, X and Y intercepts are equal, so let it be A. Then, the points are \((A, 0, 0), (0, A, 0), (0, 0, 14)\). Consider the vectors formed by any two points, say by points \((A, 0, 0)\) and \((0, A, 0)\), we get the vector \(\mathbf{u} = \langle -A, A, 0 \rangle\). And, for points \((A, 0, 0)\) and \((0, 0, 14)\), we get the vector \(\mathbf{v} = \langle -A, 0, 14 \rangle\). Calculate the cross product of these vectors to find the normal vector \(\mathbf{n}\).

To find the normal vector, we need to compute the cross product of \(\mathbf{u}\) and \(\mathbf{v}\): \[\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{bmatrix} i & j & k \\ -A & A & 0 \\ -A & 0 & 14 \end{bmatrix}\] Expanding the determinant, we get: \[\mathbf{n} = \langle 14A, -14A, -A^2 \rangle\]

Using the point-normal form of the plane equation: \[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\] Plug in the point \((2,1,3)\) as \((x_0, y_0, z_0)\) and the components of the normal vector, \(\langle 14A, -14A, -A^2 \rangle\), as \((A, B, C)\): \[14A(x - 2) - 14A (y - 1) - A^2(z - 3) = 0\]

Simplify the equation by factoring out common terms to find the answer: \[14A (x - y) - A^2(z - 3) = 0\] Now let's divide the equation by A: \[14(x - y) - Az = 42\] Comparing this equation to the given choices, the correct answer is: (A) \(11x - 11y + 3z = 42\)