The triad (x, y, z) of real numbers such that | vedclass

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(C) Given the vector equation: $(3 \hat{i}-\hat{j}+2 \hat{k}) = x(2 \hat{i}+3 \hat{j}-\hat{k}) + y(\hat{i}-2 \hat{j}+2 \hat{k}) + z(-2 \hat{i}+\hat{j}

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$(2, 5, 3)$

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(C) Given the vector equation: $(3 \hat{i}-\hat{j}+2 \hat{k}) = x(2 \hat{i}+3 \hat{j}-\hat{k}) + y(\hat{i}-2 \hat{j}+2 \hat{k}) + z(-2 \hat{i}+\hat{j}-2 \hat{k})$ Equating the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ on both sides,we obtain the following system of linear equations: $2x + y - 2z = 3$  $(1)$ $3x - 2y + z = -1$  $(2)$ $-x + 2y - 2z = 2$  $(3)$ Subtracting equation $(3)$ from equation $(1)$: $(2x + y - 2z) - (-x + 2y - 2z) = 3 - 2$ $3x - y = 1 \implies y = 3x - 1$ Substitute $y = 3x - 1$ into equation $(2)$: $3x - 2(3x - 1) + z = -1$ $3x - 6x + 2 + z = -1 \implies z = 3x - 3$ Substitute $y$ and $z$ into equation $(1)$: $2x + (3x - 1) - 2(3x - 3) = 3$ $2x + 3x - 1 - 6x + 6 = 3$ $-x + 5 = 3 \implies x = 2$ Now,find $y$ and $z$: $y = 3(2) - 1 = 5$ $z = 3(2) - 3 = 3$ Thus,the triad $(x, y, z)$ is $(2, 5, 3)$.

The triad $(x, y, z)$ of real numbers such that $(3 \hat{i}-\hat{j}+2 \hat{k})=(2 \hat{i}+3 \hat{j}-\hat{k}) x+(\hat{i}-2 \hat{j}+2 \hat{k}) y+(-2 \hat{i}+\hat{j}-2 \hat{k}) z$ is

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