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Step 1: Express the vector \vec{x} in the plane spanned by \vec{a} and \vec{b}
Since \vec{x} lies in the plane containing vectors \vec{a} and \vec{b} , we can write:
\vec{x} = k \bigl(\vec{a} + \lambda \,\vec{b}\bigr) ,
where k and \lambda are real constants to be determined.
Step 2: Impose the perpendicularity condition
We are given that \vec{x} is perpendicular to the vector 3\hat{i} + 2\hat{j} - \hat{k} . Hence, their dot product must be zero:
\vec{x} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 0 \,.
Substituting \vec{x} = k(\vec{a} + \lambda\vec{b}) and simplifying gives an equation in \lambda .
Step 3: Compute the dot product to find \lambda
Let \vec{a} = 2\hat{i} - \hat{j} + \hat{k} and \vec{b} = \hat{i} + 2\hat{j} - \hat{k} . Then
\vec{a} + \lambda \vec{b} = (2 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (1 - \lambda)\hat{k}.
The dot product with (3\hat{i} + 2\hat{j} - \hat{k}) is:
[(2 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (1 - \lambda)\hat{k}] \,\cdot\, (3\hat{i} + 2\hat{j} - \hat{k}) = 0.
Expanding and multiplying by k (since k is a common scalar factor) leads to an equation in \lambda . Solving it yields:
\lambda = -\frac{3}{8} \,.
Step 4: Use the projection condition
The projection of \vec{x} on \vec{a} is given to be \tfrac{17\sqrt{6}}{2} . Recall that the projection of a vector \vec{u} on \vec{v} is
\dfrac{\vec{u}\cdot \vec{v}}{\lVert \vec{v}\rVert}.
Hence,
\frac{\vec{x} \cdot \vec{a}}{\lvert \vec{a}\rvert} = \frac{17\sqrt{6}}{2}.
We know \vec{a} = 2\hat{i} - \hat{j} + \hat{k} , so |\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}.
Substitute \vec{x} = k\bigl(\vec{a} + \lambda \vec{b}\bigr) with \lambda = -\frac{3}{8} , then solve for k .
Step 5: Determine k and find \vec{x} explicitly
Carrying out the dot product and equating it to the correct projection value leads to
k=8.
Substitute k=8 and \lambda = -\frac{3}{8} back into
\vec{x} = k (\vec{a} + \lambda \vec{b}):
\vec{x} = 8 \,\Bigl(\vec{a} + \bigl(-\tfrac{3}{8}\bigr)\vec{b}\Bigr).
Upon simplifying, we get:
\vec{x} = 13\hat{i} \;-\; 14\hat{j} \;+\; 11\hat{k}.
Step 6: Compute \lvert \vec{x}\rvert^2
The square of the magnitude is:
\lvert \vec{x}\rvert^2
= (13)^2 + (-14)^2 + (11)^2
= 169 + 196 + 121
= 486.
Thus, the required value of \lvert \vec{x}\rvert^2 is \boxed{486}.
