Do you mean to enter (1, -1, -5)...
No. Why would that make any sense to do? How would plugging in \(x_1 = 1, \: x_2 = -1, \: x_3 = -5\) give you any insight into whether or not \(x_1 = 3, \: x_2 = 1, \: x_3 = 1\) is a solution?
What Dr. P. meant was to check to make sure you copied down the values correctly (hint: you didn't). Let \(x_1 = {\color{red}3}, \: x_2 = 1, \: x_3 = 1\) and plug in:
\(\displaystyle
2({\color{red}3}) - 4(1) - (1) = 1 \: \: \: \checkmark \\
({\color{red}3}) - 3(1) + (1) = 1 \: \: \: \checkmark \\
3({\color{red}3}) - 5(1) - 3(1) = 1 \: \: \: \checkmark
\)
Contrast that with what you wrote:
\(\displaystyle
2({\color{red}3}) - 4(1) - (1) = 1 \\
({\color{red}1}) - 3(1) + (1) = -1 \\
3({\color{red}1}) - 5(1) - 3(1) = -5
\)