Answer
$1$
Work Step by Step
Step 1. Start from the original, use row operations indicated on the right side of the matrix:
$LHS=\begin{vmatrix} x&1&1\\1&1&x\\x&1&x \end{vmatrix}\begin{array}( \\R_2-R_1\to R_2 \\R_3-R_1\to R_3 \\ \end{array}$
$LHS=\begin{vmatrix} x&1&1\\1-x&0&x-1\\0&0&x-1 \end{vmatrix} $
Step 2. Use row3 to expand and get: $LHS=(x-1)\begin{vmatrix} x&1\\1-x&0 \end{vmatrix} $
Step 3. Calculate the last determinant to get: $(x-1)(0-(1-x))=0$ or $(x-1)^2=0$
Step 4. The solution to the equation above gives $x=1$
