If (x, y, z) ne (0, 0, 0) and (i + j + 3k)x + | vedclass

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(D) Given the vector equation: $(x + 3y - 4z)i + (x - 3y + 5z)j + (3x + y)k = \lambda xi + \lambda yj + \lambda zk$. Comparing the coefficients of $i,

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$0, -1$

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(D) Given the vector equation: $(x + 3y - 4z)i + (x - 3y + 5z)j + (3x + y)k = \lambda xi + \lambda yj + \lambda zk$. Comparing the coefficients of $i, j,$ and $k$ on both sides, we get the following system of linear equations: $(1 - \lambda)x + 3y - 4z = 0$ ... $(i)$ $x - (3 + \lambda)y + 5z = 0$ ... $(ii)$ $3x + y - \lambda z = 0$ ... $(iii)$ Since $(x, y, z) 
e (0, 0, 0)$, the system has a non-trivial solution, which implies the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 1 - \lambda & 3 & -4 \ 1 & -(3 + \lambda) & 5 \ 3 & 1 & -\lambda \end{vmatrix} = 0$. Expanding the determinant: $(1 - \lambda) [\lambda(3 + \lambda) - 5] - 3 [-\lambda - 15] - 4 [1 + 3(3 + \lambda)] = 0$ $(1 - \lambda) [\lambda^2 + 3\lambda - 5] + 3\lambda + 45 - 4 [1 + 9 + 3\lambda] = 0$ $(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$ $-\lambda^3 - 2\lambda^2 - \lambda = 0$ $-\lambda(\lambda^2 + 2\lambda + 1) = 0$ $-\lambda(\lambda + 1)^2 = 0$. Wait, re-evaluating the determinant expansion: $(1 - \lambda)(\lambda^2 + 3\lambda - 5) - 3(-\lambda - 15) - 4(1 + 9 + 3\lambda) = 0$ $(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$ $-\lambda^3 - 2\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda^2 + 2\lambda + 1) = 0 \Rightarrow \lambda(\lambda + 1)^2 = 0$. Checking the original equations again: $x + 3y - 4z = \lambda x \Rightarrow (1-\lambda)x + 3y - 4z = 0$ $x - 3y + 5z = \lambda y \Rightarrow x - (3+\lambda)y + 5z = 0$ $3x + y = \lambda z \Rightarrow 3x + y - \lambda z = 0$ Determinant: $(1-\lambda)(\lambda^2+3\lambda-5) - 3(-\lambda-15) - 4(1+9+3\lambda) = 0$ $-\lambda^3 - 2\lambda^2 - \lambda + 0 = 0 \Rightarrow \lambda = 0, -1$. Thus, the values are $0, -1$.

If $(x, y, z) \ne (0, 0, 0)$ and $(i + j + 3k)x + (3i - 3j + k)y + (-4i + 5j)z = \lambda (xi + yj + zk),$ then the value of $\lambda$ will be

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