Let $\overrightarrow{\mathrm{X}}$ be a vector in the plane containing vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$. If the vector $\overrightarrow{\mathrm{x}}$ is perpendicular to $(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$ and its projection on $\overrightarrow{\mathrm{a}}$ is $\frac{17 \sqrt{6}}{2}$, then the value of $|\vec{x}|^{2}$ is equal to
Let $\overrightarrow{\mathrm{x}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}}$ ( $\lambda$ and $\mu$ are scalars)
$\overrightarrow{\mathrm{x}}=\hat{\mathrm{i}}(2 \lambda+\mu)+\hat{\mathrm{j}}(2 \mu-\lambda)+\hat{\mathrm{k}}(\lambda-\mu)$
Since $\overrightarrow{\mathrm{x}} \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0$
$3 \lambda+8 \mu=0$..(1)
Also Projection of $\vec{x}$ on $\vec{a}$ is $\frac{17 \sqrt{6}}{2}$
$\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|}=\frac{17 \sqrt{6}}{2}$
$6 \lambda-\mu=51$..(2)
From $(1)$ and $(2)$
$\lambda=8, \mu=-3$
$\overrightarrow{\mathrm{x}}=13 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$
$|\overrightarrow{\mathrm{x}}|^{2}=486$
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